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Parameterized complexity of abduction in Schaefer’s framework

Parameterized complexity of abduction in Schaefer’s framework Abstract Abductive reasoning is a non-monotonic formalism stemming from the work of Peirce. It describes the process of deriving the most plausible explanations of known facts. Considering the positive version, asking for sets of variables as explanations, we study, besides the problem of wether there exists a set of explanations, two explanation size limited variants of this reasoning problem (less than or equal to, and equal to a given size bound). In this paper, we present a thorough two-dimensional classification of these problems: the first dimension is regarding the parameterized complexity under a wealth of different parameterizations, and the second dimension spans through all possible Boolean fragments of these problems in Schaefer’s constraint satisfaction framework with co-clones (T. J. Schaefer. The complexity of satisfiability problems. In Proceedings of the 10th Annual ACM Symposium on Theory of Computing, May 1–3, 1978, San Diego, California, USA, R.J. Lipton, W.A. Burkhard, W.J. Savitch, E.P. Friedman, A.V. Aho eds, pp. 216–226. ACM, 1978). Thereby, we almost complete the parameterized complexity classification program initiated by Fellows et al. (The parameterized complexity of abduction. In Proceedings of the Twenty-Sixth AAAI Conference on Articial Intelligence, July 22–26, 2012, Toronto, Ontario, Canada, J. Homann, B. Selman eds. AAAI Press, 2012), partially building on the results by Nordh and Zanuttini (What makes propositional abduction tractable. Artificial Intelligence, 172, 1245–1284, 2008). In this process, we outline a fine-grained analysis of the inherent parameterized intractability of these problems and pinpoint their FPT parts. As the standard algebraic approach is not applicable to our problems, we develop an alternative method that makes the algebraic tools partially available again. 1 Introduction The framework of parameterized complexity theory yields a more fine-grained complexity analysis of problems than the classical worst-case complexity may achieve. Introduced by Downey and Fellows [17, 18], one associates problems with a specific parameterization, i.e. one studies the complexity of parameterized problems. Here, one aims to find parameters relevant for practice allowing to solve the problem by algorithms running in time |$f(k)\cdot n^{O(1)}$|⁠, where |$f$| is a computable function, |$k$| is the value of the parameter and |$n$| is the input length. Problems with such a running time are called fixed-parameter tractable (⁠|${\textbf{FPT}}$|⁠) and correspond to efficient computation in the parameterized setting. This is justified by the fact that parameters are usually slowly growing or even of constant value. Despite that, a different quality of runtimes is of the form |$n^{f(k)}$|⁠, which are obeyed by algorithms solving problems in the class |${\textbf{XP}}$|⁠. Comparing both classes with respect to the runtimes their problems allow to be solved in, of course, both runtimes are polynomial. However, for the first type, the degree of the polynomial is independent of the parameter’s value which is notable to observe. As a result, the second kind of runtimes is undesirable and usually tried to circumvented by locating different parameters. It is known that |${\textbf{FPT}}\subsetneq{\textbf{XP}}$| by diagonalization and also that a (presumably infinite) hierarchy of parameterized intractability in between these two classes exist: the so-called |$\textbf{W}$|-hierarchy, which is contained also in the class |$\textbf{W}[\textbf{P}] \subseteq{\textbf{XP}}$|⁠. These |$\textbf{W}$|-classes are regarded as a measure of intractability in the parameterized sense. Intuitively, showing |$\textbf{W}[1]$|-lower bounds corresponds to |${\textbf{NP}} $|-lower bounds in the classical setting. The limit of this hierarchy, the class |$\textbf{W}[\textbf{P}] $| is defined via nondeterministic machines that have at most |$h(k)\cdot \log n$| many nondeterministic steps, where |$h$| is a computable function, |$k$| the parameter’s value and |$n$| is the input length. Clearly, the process of human common-sense reasoning is non-monotonic, as adding further knowledge might decrease the number of deducible facts. As a result, non-monotonic logics became a well-established approach to investigate this kind of reasoning. One of the popular formalism in this area of research is abductive reasoning, which is an important concept in artificial intelligence as emphasized by Morgan [30] and Pole [36]. In particular, abduction is used in the process of medical diagnosis [24, 35] and thereby relevant for practice. Intuitively, abductive reasoning describes the process of deriving the most plausible explanations of known facts and originated from the work by Peirce [34]. Formally, one uses propositional formulas to model known facts in a knowledge base|${\textit{KB}}$| together with a set of manifestations|$M$| and a set of hypotheses|$H$|⁠. In this paper, |$H$| and |$M$| are sets of propositions as studied by Fellows et al. [21] as well as Eiter and Gottlob [20]. Formally, one tries to find a preferably small set of propositions |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|⁠. |$E$| is then called an explanation for |$M$|⁠. In this context, we distinguish three kinds of problems: the first just asks for such a very set |$E$| that fulfils these properties (⁠|$\textrm{ABD}$|⁠), the second tries to find a set of size less than or equal to a specific size (⁠|$\textrm{ABD}_\leq $|⁠) and the third one wants to spot a set of exactly a given size (⁠|$\textrm{ABD}_=$|⁠). Classically, |$\textrm{ABD}$| is complete for the second level of the polynomial hierarchy |${\boldsymbol{\varSigma _2^P}} $| [20] and its difficulty is very well understood [11, 16, 31, 42]. As a result, under reasonable complexity-theoretic assumptions, the problem is highly intractable posing the question in turn for sources of this complexity. In this direction, there exists research that aims to better understand the structure and difficulty of this problem, i.e. in the context of parameterized complexity. Here, Fellows et al. [21] initiated an investigation of possible parameters and classified CNF-induced fragments of the reasoning problems with respect to a multitude of parameters. The authors study the CNF-fragments Horn, Krom and DefHorn. They studied the parameterizations |$|M|$| (number of manifestations), |$|H|$| (number of hypotheses), |$|V|$| (number of variables) and |$|E|$| (number of explanations that is equivalent to their solution size |$k$|⁠) directly stemming from problem components, as well as the tree-width [39], and the size of the smallest vertex cover. In their classification, besides showing several |$ {\textbf{para-}}{\textbf{NP}} $|-/|$\textbf{W}[\textbf{P}]$|-complete/|$ {\textbf{FPT}} $| cases, they also focus on the existence of polynomial kernels and present a complete picture regarding their CNF-classes. Universal algebra yields a systematic way to rigorously classify fragments of a problem induced by restricting its Boolean connectives. This technique is built around Post’s lattice [37] which bases on the notion of (co-)clones. Intuitively, given a set of Boolean functions |$B$|⁠, the clone of |$B$| is the set of functions that are expressible by compositions of functions from |$B$| (plus introducing fictive variables). The most prominent result under this approach is the dichotomy theorem of Lewis [26], which classifies propositional satisfiability into polynomial-time solvable cases and intractable ones depending merely on the existence of specific Boolean operators. This approach has been followed many times in a wealth of different contexts [2, 3, 9, 14, 28, 29, 38] as well as in the context of abduction itself [12, 31]. Interestingly, in the scope of constraint satisfaction problems, the investigation of co-clones (or relational clones) allows one to proceed a similar kind of classification (see, e.g. the work by Nordh and Zanuttini [31]). The reason for that lies in the concept of invariance of relations under some function |$f$| (one defines this property via polymorphisms where |$f$| is applied component-wisely to the columns of the relation). In view of this, Post’s lattice supplies a similar lattice, now for sets of relations that are invariant under respective functions. With respect to constraint satisfaction, the most prominent classification is due to Schaefer [40] who similarly divides the constraint satisfaction problem restricted to co-clones into polynomial-time solvable and |$ {\textbf{NP}} $|-complete cases. The algebraic approach has been successfully applied to abduction by Nord and Zanuttini [31]. For the problems that we consider, it is less obvious how to use the algebraic tools: the standard trick to obtain reductions preserves the existence of explanations, but not their size. Due to this, we develop an alternative method that makes the algebraic tools partially available again (see Section 3.1). Much in the vein of Schaefer’s classification, we present a thorough study directly pinpointing those restrictions of the abductive reasoning problem, which yield efficiency under the parameterized approach. In a sense, we present an almost complete picture that has been initiated by Fellows et al. [21] except for some minor cases around the affine co-clones. Their classification is covered by our study now, as Horn cases correspond to the co-clones below |$ {\textsf{IE}}_2$|⁠, DefHorn conforms |$ {\textsf{IE}}_1$| and Krom matches with |${\textsf{ID}}_2$|⁠. The motivation of our research is to draw a finer line than Fellow et al. did and to present complete picture with respect to all possible constraint languages now. From this classification, we draw some surprising results. Regarding the essentially negative cases for the parameter |$|M|$|⁠, |$\textrm{ABD}_=$| is |${\textbf{para-}}{\textbf{NP}} $|-complete, whereas |$\textrm{ABD}_\leq $| is |${\textbf{FPT}} $|⁠. Also for this parameter, |${\textsf{IE}}_1$| and |${\textsf{IE}}$| are hard for |$\textrm{ABD}_=$| and |$\textrm{ABD}_\leq $| (both |${\textbf{para-}}{\textbf{NP}} $|-complete), but |$\textrm{ABD}$| is |${\textbf{FPT}} $|⁠. Regarding |$|E|$| as parameterization, the behaviour is similarly unexpected for the essentially negative cases: |${\textbf{FPT}} $| for |$\textrm{ABD}_\leq $| versus |$\textbf{W}[1]$|-hardness for |$\textrm{ABD}_=$|⁠. Another interesting remark regarding the essentially negative languages is that the problem |$\textrm{ABD}_={(|M|)}$| is harder than |$\textrm{ABD}_\leq{(|E|)}$| (⁠|${\textbf{para-}}{\textbf{NP}} $| vs |$\textbf{W} [2]$|⁠), whereas |$\textrm{ABD}_\leq{(|M|)}$| is easier than |$\textrm{ABD}_\leq{(|E|)}$| (⁠|$\textbf{W}[1]$| vs |$\textbf{W}[2]$|⁠). For the parameters |$|V|$| as well as |$|H|$| the classifications for all three problems are the same. Figure 1 shows our results for all problems and parameterizations in a single picture. Figure 1. Open in new tabDownload slide Complexity landscape of abductive reasoning with respect to the studied parameters |$|M|,|H|,|V|,|E|$|⁠. White colouring means unclassified. |$\textrm{ABD}_\star $| means same result for all three variants. Figure 1. Open in new tabDownload slide Complexity landscape of abductive reasoning with respect to the studied parameters |$|M|,|H|,|V|,|E|$|⁠. White colouring means unclassified. |$\textrm{ABD}_\star $| means same result for all three variants. The paper is organized as follows. We first recall the basic concepts from parameterized complexity theory, the notion of co-clones and formalize the abduction problem. Section 3 includes our complexity results, we first prove an important result in Section 3.1. Then we give some general intractable as well as tractable cases that are true under every parameterization. This is followed by subsections discussing individual parameters with Theorems summarizing our results. We conclude in Section 4 with some interesting remarks and open cases. 2 Preliminaries We require standard notions from classical complexity theory [33]. We encounter the classical complexity classes P, NP, |${\textbf{DP}} = \{A \setminus B \mid A,B\in{\textbf{NP}} \}$|⁠, |${\textbf{co}}{\textbf{NP}} $|⁠, |${\boldsymbol{\varSigma _2^P}} ={\textbf{NP}} ^{\textbf{NP}} $| and their respective completeness notions, employing polynomial time many-one reductions (⁠|$\leq ^{\textbf{P}}_m$|⁠). Parameterized Complexity Theory. A parameterized problem (PP) |$P\subseteq \varSigma ^*\times \mathbb N$| is a subset of the crossproduct of an alphabet and the natural numbers. For an instance |$(x,k)\in \varSigma ^*\times \mathbb N$|⁠, |$k$| is called the (value of the) parameter. A parameterization is a polynomial-time computable function that maps a value from |$x\in \varSigma ^*$| to its corresponding |$k\in \mathbb N$|⁠. The problem |$P$| is said to be fixed-parameter tractable (or in the class |${\textbf{FPT}} $|⁠) if there exists a deterministic algorithm |$\mathcal A$| and a computable function |$f$| such that for all |$(x,k)\in \varSigma ^*\times \mathbb N$|⁠, algorithm |$\mathcal A$| correctly decides the membership of |$(x,k)\in P$| and runs in time |$f(k)\cdot |x|^{O(1)}$|⁠. The problem |$P$| belongs to the class |${\textbf{XP}} $| if |$\mathcal A$| runs in time |$|x|^{f(k)}$|⁠. There exists a hierarchy of complexity classes in between |${\textbf{FPT}} $| and |${\textbf{XP}} $|⁠, which is called |$\textbf{W}$|-hierarchy (for details see the textbook of Flum and Grohe [22]). We will make use of the classes |$\textbf{W}[1]$| and |$\textbf{W}[2]$|⁠. Complete problems characterizing these classes are introduced later in Proposition 2.4. Also, we work with classes that can be defined via a precomputation on the parameter. Definition 2.1 Let |$\mathcal C$| be any complexity class. Then |${\textbf{para-}} \mathcal C$| is the class of all PPs |$P\subseteq \varSigma ^*\times \mathbb N$| such that there exists a computable function |$\pi \colon \mathbb N\to \varDelta ^*$| and a language |$L\in \mathcal C$| with |$L\subseteq \varSigma ^*\times \varDelta ^*$| such that for all |$(x,k)\in \varSigma ^*\times \mathbb N$| we have that |$(x,k)\in P \Leftrightarrow (x,\pi (k))\in L$|⁠. Notice that |${\textbf{para-}} \textbf{P}={\textbf{FPT}} $|⁠. The complexity classes |$\mathcal{C}\in{\{\, {\textbf{NP}} ,{\textbf{co}}{\textbf{NP}} , {\textbf{DP}} ,{\boldsymbol{\varSigma _2^P}} \,\}} $| are used in the |${\textbf{para-}} \mathcal C$| context by us. Let |$c\in \mathbb N$| and |$P\subseteq \varSigma ^*\times \mathbb N$| be a PP, then the |$c$|-slice of|$P$|⁠, written as |$P_c$| is defined as |$P_c:=\{\,(x,k)\in \varSigma ^*\times \mathbb N\mid k=c\,\}$|⁠. Notice that |$P_c$| is a classical problem then. Observe that, regarding our studied complexity classes, showing membership of a PP |$P$| in the complexity class |${\textbf{para-}} \mathcal C$|⁠, it suffices to show that each slice |$P_c\in \mathcal C$|⁠. Definition 2.2 Let |$P\subseteq \varSigma ^*\times \mathbb N$| and |$Q\subseteq \varGamma ^*\times \mathbb N$| be two PPs. One says that |$P$| is fpt-reducible to |$Q$|⁠, if there exists an fpt-computable function |$f\colon \varSigma ^*\times \mathbb N\to \varGamma ^*\times \mathbb N$| such that – for all |$(x,k)\in \varSigma ^*\times \mathbb N$| we have that |$(x,k)\in P\Leftrightarrow f(x,k)\in Q$|⁠, there exists a computable function |$g\colon \mathbb N\to \mathbb N$| such that for all |$(x,k)\in \varSigma ^*\times \mathbb N$| and |$f(x,k)=(x^{\prime},k^{\prime})$| we have that |$k^{\prime}\leq g(k)$|⁠. Propositional Logic. We assume familiarity with propositional logic. A literal is a variable |$x$| or its negation |$\neg x$|⁠. A clause is a disjunction of literals and a term is a conjunction of literals. We denote by |$\textrm{var}(\varphi )$| the variables of a formula |$\varphi $|⁠. Analogously, for a set of formulas |$F$|⁠, |$\textrm{var}(F)$| denotes |$\bigcup _{\varphi \in F}\textrm{var}(\varphi )$|⁠. We identify finite |$F$| with the conjunction of all formulas from |$F$|⁠, i.e. |$\bigwedge _{\varphi \in F} \varphi $|⁠. A mapping |$\sigma \colon \textrm{var}(\varphi ) \mapsto \{0,1\}$| is called an assignment to the variables of |$\varphi $|⁠. A model of a formula |$\varphi $| is an assignment to |$\textrm{var}(\varphi )$| that satisfies |$\varphi $|⁠. The weight of an assignment |$\sigma $| is the number of variables |$x$| such that |$\sigma (x)=1$|⁠. For two formulas |$\psi , \varphi $| we write |$\psi \models \varphi $| if every model of |$\psi $| also satisfies |$\varphi $|⁠. A formula is positive (resp., negative) if every literal appears positively (negatively) and a negation symbol appears only in front of a variable. The class of all propositional formulas is denoted by |${\textit{PROP}}$|⁠. Occasionally, in this paper, we will consider special subclasses of formulas, namely $$\begin{align*} \varGamma_{0, d} & = {\{\, \ell_1\land\ldots\land \ell_c \mid \ell_1,\ldots, \ell_c\ \textrm{are literals and}\ c\leq d \,\}},\\ \varDelta_{0, d} & = {\{\, \ell_1\lor\ldots\lor \ell_c \mid \ell_1,\ldots, \ell_c\ \textrm{are literals and}\ c\leq d \,\}},\\ \varGamma_{t, d} & = \left\{\,\bigwedge\limits_{i\in I} \alpha_i \,\middle|\, \alpha_i \in \varDelta_{t-1,d}\ \textrm{for}\ i \in I\, \right\}, \varDelta_{t, d} = \left\{\,\bigvee\limits_{i\in I} \alpha_i \,\middle|\, \alpha_i \in \varGamma_{t-1,d},\ i\ \in\ I\, \right\}. \end{align*}$$ Finally, |$\varGamma ^{+}_{t,d}$| (resp. |$\varGamma ^{-}_{t,d}$|⁠) denote the class of all positive (negative) formulas in |$\varGamma _{t,d}$|⁠. Example 2.3 Let |$\phi = \bigwedge _{i\leq m}(\neg x_{i,1} \lor \cdots \lor \neg x_{i,n_i})$| for |$1\leq n_i\leq d$| and |$d,m\in \mathbb N$|⁠. That is, |$\phi $| is a conjunction of the clauses containing negative literals. Then |$\phi \in \varGamma _{1,d}$|⁠, the so-called |$d$|-CNF. Note also that |$\phi $| is an |${{\textsf{IS}^{d}_{1}}}$|-formula using only negative clauses. We will often reduce a problem instance to (and from) a parameterized weighted satisfiability problem for propositional formulas. This problem is defined below. Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Open in new tab Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Open in new tab Two similarly defined problems are |${\textrm{p-WSAT}(\varGamma ^+_{t,1})}$| and |${\textrm{p-WSAT}(\varGamma ^-_{t,1})}$| where an instance |$\alpha $| comes from classes |$ \varGamma ^{+}_{t,1}$| (resp. |$\varGamma ^{-}_{t,1}$|⁠). The classes of the |$\textbf{W}$|-hierarchy can be defined in terms of these problems as proved by Downey and Fellows [22]. Proposition 2.4 ([22]). The following problems are |$\textbf{W}[t]$|-complete for each |$t \geq 1$|⁠, under |$\leq ^{\textbf{FPT}}$|-reductions: – |${\textrm{p-WSAT}(\varGamma ^+_{t,1})} $| if t is even, |${\textrm{p-WSAT}(\varGamma ^-_{t,1})} $| if t is odd, |${\textrm{p-WSAT}(\varGamma _{t,d})}$| for every |$t$| and |$d\geq 1$|⁠. Constraints and |${S}$|-formulas. A logical relation of arity |$k$| is a relation |$R \subseteq \{0,1\}^k$|⁠. A constraint is a formula |$R(x_1, \dots , x_k)$|⁠, where |$R$| is a logical relation of arity |$k$| and the |$x_i$|’s are (not necessarily distinct) variables. An assignment |$\sigma $| to the |$x_i$|’s satisfies the constraint if |$(\sigma (x_1), \dots , \sigma (x_k)) \in R$|⁠. A constraint language|${S}$| is a finite set of logical relations. An |${S}$|-formula|$\varphi $| is a conjunction of constraints built upon logical relations only from |${S}$| and accordingly can be seen as a quantifier-free first-order formula. An assignment |$\sigma $| is called a model of |$\varphi $| if |$\sigma $| satisfies all constraints in |$\varphi $| simultaneously. Whenever an |${S}$|-formula or constraint is logically equivalent to a single clause or term, we treat it as such. Definition 2.5 1. The set |$\left \langle{S} \right \rangle $| is the smallest set of relations that contains |${S}$|⁠, the equality constraint, =, and which is closed under primitive positive first order definitions, i.e. if |$\phi $| is an |${S} \cup \{=\}$|-formula and |$R(x_1, \dots , x_n) \equiv \exists y_1 \dots \exists y_l \phi (x_1, \dots , x_n,y_1, \dots , y_l)$|⁠, then |$R \in \left \langle{S} \right \rangle $|⁠. In other words, |$\left \langle{S} \right \rangle $| is the set of relations that can be expressed as an |${S} \cup \{=\}$|-formula with existentially quantified variables. 2. The set |$\left \langle{S} \right \rangle _{\neq }$| is the set of relations that can be expressed as an |${S}$|-formula with existentially quantified variables (no equality relation is allowed). The set |$\left \langle{S} \right \rangle $| is called a relational clone or co-clone with base|${S}$| [4]. Notice that for a co-clone |${\textsf{C}}$| and a constraint language |${S}$| the statements |${S} \subseteq{\textsf{C}}$|⁠, |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}$| and |$\left \langle{S} \right \rangle _{\neq } \subseteq{\textsf{C}}$| are equivalent. Throughout the text, we refer to different types of Boolean relations and corresponding co-clones following Schaefer’s terminology [40]. For an overview of co-clones and bases, see Table 1. Note that |$\left \langle{S} \right \rangle _{\neq } \subseteq \left \langle{S} \right \rangle $| by definition. The other direction does not hold in general. However, if |$(x = y) \in \left \langle{S} \right \rangle _{\neq }$|⁠, then |$\left \langle{S} \right \rangle _{\neq } = \left \langle{S} \right \rangle $|⁠. Table 1. Overview of bases [4] and clause descriptions [31] for co-clones, where EVEN|$^4$| = |$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus 1$|⁠. Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Open in new tab Table 1. Overview of bases [4] and clause descriptions [31] for co-clones, where EVEN|$^4$| = |$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus 1$|⁠. Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Open in new tab Abduction. An instance of the abduction problem for |${S}$|-formulas is given by |$\langle V, H, M, {\textit{KB}} \rangle $|⁠, where |$V$| is the set of variables, |$H$| is the set of hypotheses, |$M$| is the set of manifestations and |${\textit{KB}}$| is the knowledge base (or theory) built upon variables from |$V$|⁠. A knowledge base |${\textit{KB}}$| is a set of |${S}$|-formulas that we assimilate with the conjunction of all formulas it contains. We define the following abduction problems for |$S$|-formulas. Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Similarly, the problem |$\textrm{ABD}(S)$| is the classical pendant of |$\textrm{ABD}(S,k)$|⁠. Additionally, we consider size restrictions for a solution and define the following problems. Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Analogously, |$\textrm{ABD}_=(S,k)$| requires the size of |$E$| to be exactly |$s$| and |$\textrm{ABD}_=(S), \textrm{ABD}_\leq(S)$| are the classical counterparts. Notice that, for instance, in cases where the parameter is the size of solutions, then |$s=k$|⁠. Example 2.6 Sitting in a train you realize that it is still not moving even though the clock suggests it should be. You start reasoning about it. Either some door is open, the train has delayed or that engine has failed. This form of reasoning is called abductive reasoning. Having some additional information that the operator of train usually announces in case the train is delayed or engine has failed, you deduce that some door must be opened and that train will start moving soon when all the doors are closed. Formally, one is interested in an explanation for the observed event (manifestation) |${\{\, \texttt{stop} \,\}} $|⁠. The knowledge base includes the following statements: |$-$||$\neg \texttt{moving} \leftrightarrow \texttt{stop} \qquad\qquad\qquad\qquad \ \ \ \ \ \ -\texttt{trainDelayed} \rightarrow \texttt{newTime},$| |$\neg \texttt{announcement}$|⁠, |$\texttt{moving} \rightarrow \texttt{time}, \qquad\qquad\qquad\qquad\qquad-\ (\texttt{engineFailed} \lor \texttt{trainDelayed}\ \lor$| |$\texttt{engineFailed}\rightarrow \texttt{announcement},\qquad\quad \texttt{doorOpen} )\rightarrow \texttt{stop}$|⁠, Then the set of hypotheses |${\{\, \texttt{time}, \texttt{doorOpen}, \texttt{announcement} \,\}} $| has an explanation, namely, |${\{\, \texttt{doorOpen} \,\}} $|⁠. On the other hand, |${\{\, \texttt{time} \,\}} $| does not explain the event |${\{\, \texttt{stop} \,\}} $|⁠, whereas |${\{\, \texttt{announcement} \,\}} $| is not consistent with the knowledge base. Consequently, an explanation of size |$1$| exists. There also exists an explanation of size |$2$| since |${\{\, \texttt{time}, \texttt{doorOpen} \,\}} $| is consistent with |${\textit{KB}}$| and explains |$M$|⁠. Note that having the set of hypotheses |${\{\, \texttt{engineFailed}, \texttt{doorOpen} \,\}} $| facilitates only one explanation of size |$1$|⁠, namely, |${\{\, \texttt{doorOpen} \,\}} $|⁠, even though the hypotheses set has size |$2$|⁠. Let |${\textrm{SAT}}$| and |${\textrm{IMP}}$| denote the classical satisfiability and implication problems. Given a constraint language |${S}$| then an instance of |${\textrm{SAT}}({S})$| is an |${S}$|-formula |$\varphi $| and the question is whether there exists a satisfying assignment for |$\varphi $|⁠. On the other hand, an instance of |${\textrm{IMP}}({S})$| is |$(\phi ,\psi )$| such that |$\phi , \psi $| are two |${S}$|-formulas and the question is whether |$\phi \models \psi $|⁠. We have the following observation regarding the classical |${\textrm{SAT}}$| and |${\textrm{IMP}}$| problems. Proposition 2.7 ([40, 41]). Let |${S}$| be a constraint language such that |${S} \subseteq{\textsf{C}}$| where |${\textsf{C}} \in{\{\, {\textsf{ID}}_2,{\textsf{IV}}_2, {\textsf{IE}}_2, {\textsf{IL}}_2 \,\}} $|⁠. Then |${\textrm{SAT}}({S})$| and |${\textrm{IMP}}({S})$| are both in |${\textbf{P}} $|⁠. 3 Complexity results for abductive reasoning We begin by presenting a number of technical expressivity results that allow us in the sequel to prove a crucial property for the whole classification endeavour (Lemma 3.3). 3.1 Base independence The idea of lemma 3.2 is to express equality by some other construction, we need the following proposition for the proof. ([6]) . Proposition 3.1 Let |${S}$| be a constraint language. The following is true: 1. If |${S}$| is not 1-valid, not 0-valid and complementive, then |$(x\neq y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.6.1]. 2. If |${S}$| is not 1-valid, not 0-valid, and not complementive, then |$(x\land \neg y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.6.3]. 3. If |${S}$| is 1-valid, 0-valid and not trivial, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.7]. 4. If |${S}$| is 1-valid, not 0-valid and not essentially positive, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.8.1]. 5. If |${S}$| is not 1-valid, 0-valid and not essentially negative, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.8.2]. Lemma 3.2 Let |${S}$| be a constraint language. If |${S}$| is not essentially negative and not essentially positive, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| and |$\left \langle{S} \right \rangle = \left \langle{S} \right \rangle _{\neq }$|⁠. Proof. For constraint languages (relations) that are Horn (⁠|${\textsf{IE}}_2$|⁠), dualHorn (⁠|${\textsf{IV}}_2$|⁠), essentially negative (⁠|${\textsf{IS}_{12}}$|⁠) or essentially positive (⁠|${\textsf{IS}_{02}}$|⁠) we use the following characterizations by polymorphisms (see, e.g. [15]). The binary operations of conjunction, disjunction and negation are applied coordinate-wise. |$R$| is Horn if and only if |$m_1,m_2 \in R$| implies |$m_1 \land m_2 \in R$|⁠. |$R$| is dualHorn if and only if |$m_1,m_2 \in R$| implies |$m_1 \lor m_2 \in R$|⁠. |$R$| is essentially negative if and only if |$m_1,m_2,m_3 \in R$| implies |$m_1 \land (m_2 \lor \neg m_3) \in R$|⁠. |$R$| is essentially positive if and only if |$m_1,m_2,m_3 \in R$| implies |$m_1 \lor (m_2 \land \neg m_3) \in R$|⁠. In order to complete the proof of the lemma, we make a case distinction according to whether |${S}$| is 1- and/or 0-valid. 1-valid and 0-valid. This case follows immediately from Prop. 3.1, third item. 1-valid and not 0-valid. Follows immediately from Prop. 3.1, fourth item. not 1-valid and 0-valid. Follows immediately from Prop. 3.1, fifth item. not 0-valid and not 1-valid. We make another case distinction according to whether |${S}$| is Horn and/or dualHorn. not Horn and not dualHorn. It suffices to show that inequality |$(x \neq y)$| can be expressed, since |$(x=y) \equiv \exists z(x\neq z) \land (z\neq y)$|⁠. If |${S}$| is complementive, we obtain by Prop. 3.1, first item, that |$(x\neq y) \in \left \langle{S} \right \rangle _{\neq }$|⁠. Therefore suppose now that |${S}$| is not complementive.Let |$R$| be a relation that is not Horn. Then there are |$m_1,m_2 \in R$| such that |$m_1 \land m_2 \notin R$|⁠. For |$i,j \in \{0,1\}$|⁠, set |$V_{i,j} = \{x \mid x \in V,\ m_1(x) = i,\ m_2(x) = j\}$|⁠. Observe that the sets |$V_{0,1}$| and |$V_{1,0}$| are nonempty (otherwise |$m_1 = m_1 \land m_2$| or |$m_2 = m_1 \land m_2$|⁠, a contradiction). Denote by |$C$| the |$\{R\}$|-constraint |$C = R(x_1, \dots , x_k)$|⁠. Set $$\begin{align*} &M_1(u,x,y,v) = C[V_{0,0}/u, V_{0,1}/x, V_{1,0}/y, V_{1,1}/v].\end{align*}$$ It contains |$\{0011, 0101\}$| (since |$m_1, m_2 \in R$|⁠), but it does not contain |$0001$| (since |$m_1 \land m_2 \notin R$|⁠). Let |$R$| be a relation that is not dualHorn. Then there are |$m_3,m_4 \in R$| such that |$m_3 \lor m_4 \notin R$|⁠. For |$i,j \in \{0,1\}$|⁠, set |$V^{\prime}_{i,j} = \{x \mid x \in V,\ m_3(x) = i,\ m_4(x) = j\}$|⁠. Observe that the sets |$V^{\prime}_{0,1}$| and |$V^{\prime}_{1,0}$| are nonempty (otherwise |$m_3 = m_3 \lor m_4$| or |$m_4 = m_3 \lor m_4$|⁠, a contradiction). Set |$M_2(u,x,y,v) = C[V^{\prime}_{0,0}/u, V^{\prime}_{0,1}/x, V^{\prime}_{1,0}/y, V^{\prime}_{1,1}/v]$|⁠. It contains |$\{0011, 0101\}$| (since |$m_3, m_4 \in R$|⁠), but it does not contain |$0111$| (since |$m_3 \lor m_4 \notin R$|⁠). Finally consider the |$\{R, (t \land \neg f)\}$|-formula $$\begin{align*} &M(f,x,y,t) = M_1(f,x,y,t) \land M_2(f,x,y,t) \land (t \land \neg f).\end{align*}$$ One verifies that it is equivalent to |$(x\neq y) \land (t \land \neg f)$|⁠. Due to Prop. 3.1, second item, |$(t \land \neg f)$| is expressible as an |$S$|-formula, and therefore so is |$M(f,x,y,t)$|⁠. Since |$\exists t,f\,\! M(f,x,y,t)$| is equivalent to |$(x\neq y)$|⁠, we obtain |$(x\neq y) \in \left \langle{S} \right \rangle _{\neq }$|⁠. Horn. Let |$R$| be a relation that is not essentially negative, but Horn. Then there are |$m_1,m_2,m_3 \in R$| such that |$m_4:= m_1 \land (m_2 \lor \neg m_3) \notin R$|⁠. Since |$R$| is Horn, |$m_5:= m_1 \land m_2 \in R$|⁠. For |$i,j,k \in \{0,1\}$|⁠, set |$V_{i,j,k} = \{x \mid x \in V,\ m_1(x) = i,\ m_2(x) = j,\ m_3(x) = k\}$|⁠. Observe that the sets |$V_{1,0,0}$| and |$V_{1,0,1}$| are nonempty (otherwise |$m_5 = m_4$| or |$m_1 = m_4$|⁠, a contradiction). Denote by |$C$| the |$\{R\}$|-constraint |$C = R(x_1, \dots , x_k)$|⁠. Set |$M(f,x,y,t) = C[V_{0,0,0}/f, V_{0,0,1}/f, V_{0,1,0}/f, V_{0,1,1}/f, V_{1,0,0}/x, V_{1,0,1}/y, V_{1,1,0}/t, V_{1,1,1}/t]$|⁠. It contains |$\{00001111,00000011\}$| (since |$m_1, m_5 \in R$|⁠), but it does not contain |$00001011$| (since |$m_4 \notin R$|⁠). Finally consider the |$\{R, (t\land \neg f)$|-formula $$\begin{align*} &M^{\prime}(f,x,y,t) = M(f,x,y,t) \land M(f,y,x,t) \land (t \land \neg f)\end{align*}$$ One verifies that it contains |$\{0111, 0001\}$| but not |$0101$| and neither |$0011$|⁠. Therefore it is equivalent to |$(x=y) \land (t \land \neg f)$|⁠. Due to Prop. 3.1, second item, |$(t \land \neg f)$| is expressible as an |$S$|-formula, and therefore so is |$M^{\prime}(f,x,y,t)$|⁠. Since |$\exists t,f\,\! M^{\prime}(f,x,y,t)$| is equivalent to |$(x = y)$|⁠, we obtain |$(x = y) \in \left \langle{S} \right \rangle _{\neq }$|⁠. dualHorn. Analogously to the Horn case, using the property that |${S}$| is not essentially positive, but dualHorn. The following property is crucial for presented results in the course of this paper. It supplies generalized upper as well as lower bounds (independence of the base of a co-clone), as long as the constraint language is not essentially negative and not essentially positive. The proof idea is to implement the previous lemma. Lemma 3.3 Let |${S}, {S}^{\prime}$| be two constraint languages such that |${S}^{\prime}$| is neither essentially positive nor essentially negative. Let |$\textrm{ABD}_* \in \{\textrm{ABD}, \textrm{ABD}_=, \textrm{ABD}_\leq\}$|⁠. If |${S} \subseteq \left \langle{S}^{\prime} \right \rangle $|⁠, then |$\textrm{ABD}_*(S) \leq^{\textbf{P}}_m \textrm{ABD}_*(S^\prime)$|⁠. Proof. We may consider |${\textit{KB}}$| as a single |${S}$|-formula. We will transform |${\textit{KB}}$| into a corresponding |${S}^{\prime}$|-formula by replacing every |${S}$|-constraint by a corresponding |${S}^{\prime}$|-formula. For this purpose we first construct a look-up table mapping any |$R \in{S}$| to an |${S}^{\prime}$|-formula |$F_R$| as follows. Since |$R \in{S} \subseteq \left \langle{S}^{\prime} \right \rangle $|⁠, and by Lemma 3.2|$\left \langle{S}^{\prime} \right \rangle = \left \langle{S}^{\prime} \right \rangle _{\neq }$|⁠, we have that |$R \in \left \langle{S}^{\prime} \right \rangle _{\neq }$|⁠. Thus, we have by definition an |${S}^{\prime}$|-formula |$\phi $| such that |$R(x_1, \dots , x_n) \equiv \exists y_1 \dots \exists y_m \phi (x_1, \dots , x_n,y_1, \dots , y_m)$|⁠, where we can assume the |$x_i$|’s and |$y_i$|’s to be |$n+m$| distinct variables. We obtain |$F_R$| by removing the existential quantifiers. Note that the computation of the look-up table takes constant time, since |$S$| is finite and not dependent on any input. We are now ready to transform |${\textit{KB}}$| into an appropriate |${S}^{\prime}$|-formula by applying the following replacement procedure as long as applicable. – Let |$C_R = R(x_1, \dots , x_n)$| be an |${S}$|-constraint (now the |$x_i$|’s are not necessarily |$n$| distinct variables). Replace |$C_R$| by its corresponding |${S}^{\prime}$|-formula |$F_R(x_1, \dots , x_n, y_1, \dots , y_m)$|⁠, where the variables |$y_1, \dots , y_m$| are fresh variables that are unique to |$C_R$| (they will not be used for any other constraint replacement). This transformation procedure introduces additional variables. We show that their total number is polynomially bounded. Denote by |$m_R$| the number of |$y_i$|’s added while replacing |$C_R$| (denoted |$m$| in the above procedure). One observes that the total number of additional variables is bounded by the number of original constraints times the maximum of all |$m_R$|⁠. Since |$m_R$| is only dependent on |$R$|⁠, it is constant. Since |$S$| is finite, the maximum of all |$m_R$| is constant. We conclude that the transformation can be achieved in polynomial time. Furthermore, observe that the so obtained abduction instance has exactly the same solutions as the original instance. The last lemma in this section takes care of the essentially positive cases. The proof idea is to remove the equality clauses maintaining the size counts and the satisfiability property. Lemma 3.4 Let |${S}, {S}^{\prime}$| be two constraint languages such that |${S}^{\prime}$| is essentially positive. Let |$\textrm{ABD}_* \in \{\textrm{ABD}, \textrm{ABD}_=, \textrm{ABD}_\leq\}$|⁠. If |${S} \subseteq \left \langle{S}^{\prime} \right \rangle $|⁠, then |$\textrm{ABD}_*(S) \leq^{\textbf{P}}_m \textrm{ABD}_*(S^\prime)$|⁠. Proof. The general case is due to Nordh and Zanuttini [31, Lemma 22]. The result for ‘|$\leq $|’ is because of the following. Removing equality clauses and deleting the duplicating occurrences of variables only decrease the size of |$H$|⁠. Now we proceed with proving the case for ‘=’. We show that for any |$\textrm{ABD}_=(S \cup \{=\})$| instance |${(V, H, M, {\textit{KB}},s)}{}$|⁠, there is an |$\textrm{ABD}_=(S)$|-instance |${(V_1, H_1, M_1, {\textit{KB}}_1,s)}{}$| such that the former has an explanation if and only if the later has one. The proof uses the fact that the only negative clauses in |${\textit{KB}}$| are of size |$1$|⁠. Since the existence of a solution is invariant under the equality clauses, we only need to assure that the size of a solution is also preserved. For each clause |$x_i= x_j \in{\textit{KB}}$|⁠, we do the following: If at most one of the |$x_i,x_j$| appears in |$H$|⁠, remove the clause |$x_i = x_j$| from |${\textit{KB}}$|⁠, replace |$x_j$| by |$x_i$| everywhere in |${\textit{KB}} \cup H\cup V \cup M$| (and delete |$x_j$|⁠). If both |$x_i, x_j$| are from |$H$|⁠. Then, – if |$\neg x_i$| (resp., |$\neg x_j$|⁠) appears in |${\textit{KB}}$|⁠, we add |$\neg x_j$| (resp., |$\neg x_i$|⁠) to |${\textit{KB}}$| and remove the clause |$x_i = x_j$| from |${\textit{KB}}$|⁠. otherwise |$\neg x_i,\neg x_j\notin{\textit{KB}}$| and then simply remove the clause |$x_i = x_j$| from |${\textit{KB}}$| and do not remove any variable. The problem caused by equality clauses is the following. If we remove a variable that is also a hypothesis, then removing this from |$H$|⁠, owing to some equality constraint, may not preserve the size of solutions. Furthermore, this problem occurs only when an equality clause contains both variables from |$H$| (case 2.), since otherwise the size of |$H$| is not changed (case 1.). We prove the following correspondence between the solutions of the two instances. Claim. A subset |$E\subseteq H$| is an explanation for |$\textrm{ABD}_=(S \cup \{=\})$| if and only if |$E$| is an explanation for |$\textrm{ABD}_=(S)$|⁠. Proof of Claim. ‘|$\!\!\!\!\implies\!\!\!\!$|’: Let |${(V,H,M,{\textit{KB}},s)}$| be an instance of |$\textrm{ABD}_=(S \cup \{=\})$| and let |${(V_1,H,M_1,{\textit{KB}}_1,s)}$| be the corresponding instance of |$\textrm{ABD}_=(S \cup \{=\})$|⁠, where |${\textit{KB}}_1$| is obtained from |${\textit{KB}}$| by applying steps 1 and 2 above. If |$E\land{\textit{KB}}$| is consistent, we prove that |$E\land{\textit{KB}}_1$| is also consistent. Note that |${\textit{KB}}_1 \subseteq{\textit{KB}}$|⁠, except if |$\neg x_j\in{\textit{KB}}_1$| for some |$x_j$|⁠. This implies that |$x_j\not \in E$| due to the reason that |$x_i = x_j$| and |$\neg x_i \in{\textit{KB}}$|⁠. Finally, |$M_1\subseteq M$| and |$E$| is an explanation for |$M$| implies |$E$| is also an explanation for |$M_1$|⁠. ‘|$\!\!\impliedby\! $|’: Suppose that |$E \land{\textit{KB}}_1$| is consistent and let |$\theta $| be a satisfying assignment. We consider each equality constraint separately and prove that |$E \land{\textit{KB}}$| is consistent. In the first case, for each |$x_i, x_j$| such that at most one (say |$x_i$|⁠) appears in |$H$|⁠. If |$x_i \in E \subseteq H$| then |$\neg x_j \not \in{\textit{KB}}_1$| since this would imply that |$\neg x_i \in{\textit{KB}}_1$| and |$x_i \not \in E$|⁠. Consequently, |$E \land{\textit{KB}}_1 \land (x_i=x_j)$| is consistent (by extending |$\theta $| to |$\theta (x_i)= 1 =\theta (x_j)$|⁠). On the other hand, if |$\neg x_j \in{\textit{KB}}_1$| then |$\neg x_i \in{\textit{KB}}_1$| and |$x_i \not \in E$|⁠. As a result, |$\theta $| extended to |$\theta (x_i)= 0 =\theta (x_j)$| is a satisfying assignment. In the second case, if both |$x_i, x_j \in H$| then we also have two sub-cases based on whether |$\neg x_i \in{\textit{KB}}_1$| or not. If |$\neg x_i \in{\textit{KB}}_1$| then due to case 2, we have |$\neg x_j\in{\textit{KB}}_1$| and this implies |$x_i\not \in E$|⁠. As a consequence, |$E \land{\textit{KB}}_1 \land (x_i=x_j)$| is consistent by extending |$\theta $| to |$\theta (x_i)= 0 =\theta (x_j)$|⁠. In the sub-case when both |$x_i, x_j \in H$| and |$\neg x_j \not \in{\textit{KB}}_1$| then mapping |$\theta (x_i)=1 = \theta (x_j)$| satisfies |$E\land{\textit{KB}}_1 \land (x_i= x_j)$|⁠. This is because all the non-unit clauses in |${\textit{KB}}_1$| are positive. Since this is true for all the equality clauses it shows that |$E\land{\textit{KB}}$| is consistent. For entailment, let |$m_i, m_j \in M$||$(m_i = m_j) \in{\textit{KB}}$| and |$m_j \not \in M_1$|⁠, i.e. |$M_1\subsetneq M$|⁠. Since |$E\land{\textit{KB}}_1$| is consistent and entails |$M_1$|⁠, we have |$E\land{\textit{KB}}_1 \land (m_i=m_j)$| is also consistent (due to arguments for consistency) and entails |$M_1 \cup \{m_j\}$|⁠. This completes the proof in this direction and settles the claim. Finally, the above reduction can be computed in polynomial time because both steps are applied once for each equality clause and each step takes polynomial time. This shows the desired reduction between |$\textrm{ABD}_=(S \cup \{=\})$| and |$\textrm{ABD}_=(S)$|⁠. REMARK 3.5 Notice that Lemmas 3.3 and 3.4 are stated with respect to the classical and unparameterized decision problems. However, these reductions can be generalized to |$\leq ^{\textbf{FPT}}$|-reductions whenever the parameters are bound as required by Def. 2.2. That is, in our case, for any parameterization |$k\in{\{\, |H|,|E|,|M| \,\}} $| the reductions are valid. Even more, the values of the parameters stay the same as in the reduction the sizes of |$H$|⁠, |$E$| and |$M$| remain unchanged. REMARK 3.6 It is rather cumbersome to mention the base independence results in almost every single proof. As a result, we omit this reference and show the results only for concrete bases, thereby, implicitly using the above lemmas. In cases where we deal with essentially negative constraint languages, we do not have a general base independence result but direct constructions showing membership and hardness in our cases for all bases (e.g. Lemmas 3.13 and 3.23). 3.2 General complexity results In this section, we start with general observations and reductions between the defined problems. Then we prove some immediate (parameterized) complexity results. We provide two results that help us to consider fewer cases to solve. Lemma 3.7 For every constraint language |$S$| we have |$\textrm{ABD}(S) \leq^{\textbf{P}}_m \textrm{ABD}_\leq(S)$|⁠. Proof. Clearly, |${(V,H,M,KB)} \in \textrm{ABD}(S) \Leftrightarrow{(V,H,M,KB,s)} \in \textrm{ABD}_\leq(S)$|⁠, where |$s=|H|$|⁠. That is, there is an explanation for an abduction instance if and only if there is one with size at most that of the hypotheses set. Lemma 3.8 |$\textrm{ABD}_\leq(S) = \textrm{ABD}_=(S)$| for any |${S}$| such that |${\textsf{IBF}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. Proof. ‘|$\subseteq $|’: We claim that every positive instance |${(V,H,M,{\textit{KB}},s)}\in \textrm{ABD}_\leq(S)$| has a solution |$E$| of size exactly |$s$|⁠. Given a solution of size |$\leq s$| then a solution of size |$=s$| can be constructed from it (in even polynomial time w.r.t. |$|H|$|⁠) by adding one element |$h$| at a time from |$H$| to |$E$| and checking that |$\neg h \not \in{\textit{KB}}$|⁠. ‘|$\supseteq $|’: Every solution of size exactly |$s$| is a solution of size |$\le s$|⁠. Intractable cases It turns out that for |$0$|-valid, |$1$|-valid and complementive languages, all three problems remain hard under any parameterization except the case |$|V|$|⁠. Theoem 3.9 The problems |$\textrm{ABD}(S, k)$|⁠, |$\textrm{ABD}_\leq(S, k)$|⁠, |$\textrm{ABD}_=(S, k)$| are |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$| and |$k\in{\{\, |H|,|E|, |M| \,\}} $|⁠, |${\textbf{para-}}{\textbf{DP}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$| and |$k\in \{|H|,|E|\}$|⁠. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |$k=|M|$| for |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠. Proof. (1.) We prove the case for |${\textsf{IN}}$| regarding all three parameters simultaneously. Notice that |${\textrm{IMP}}({\textsf{II}}_1)$| is |${\textbf{co}}{\textbf{NP}} $|-hard [31, Thm. 34] even if the right side contains only a single variable. We describe in the following a modified proof from [31, Prop. 48]. Since |$\left \langle{\textsf{IN}} \cup{\{\, T \,\}} \right \rangle = {\textsf{II}}_1$| (define |$T(x) \equiv x$|⁠) we have that |${\textrm{IMP}}({\textsf{IN}} \cup{\{\, T \,\}} )$| is |${\textbf{co}}{\textbf{NP}} \text{-hard} $|⁠, even if the right side contains only a single variable. We reduce |${\textrm{IMP}}({\textsf{IN}} \cup{\{\, T \,\}} )$| to our abduction problems with |$|H|=1$|⁠, |$|M|=1$| and |$|E|=1$|⁠. Let |$({\textit{KB}}_T, q)$| be an instance of |${\textrm{IMP}}({\textsf{IN}} \cup{\{\, T \,\}} )$|⁠, where |${\textit{KB}}_T = {\textit{KB}} \land \bigwedge _{x\in V_T} T(x)$| with |${\textit{KB}}$| being an |${\textsf{IN}}$|-formula. We map |$({\textit{KB}}_T, q)$| to |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime})}$|⁠, where |$V = \textrm{var}({\textit{KB}}) \cup \{h\}$|⁠, |$h$| is a fresh variable and |${\textit{KB}}^{\prime}$| is obtained from |${\textit{KB}}$| by replacing any variable from |$V_T$| by |$h$|⁠. Note that |${\textit{KB}}_T \equiv{\textit{KB}}^{\prime} \land h$|⁠. Since |${\textit{KB}}$| and |${\textit{KB}}^{\prime}$| are 1-valid, clearly, |${\textit{KB}}^{\prime} \land h$| is always satisfiable and there exists an explanation iff |${\textit{KB}}^{\prime} \land h \models q$|⁠, iff |${\textit{KB}}_T \models q$|⁠. Furthermore, observe that |${\textit{KB}}_T\models q$| if and only if |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime}, |H|)} \in \textrm{ABD}\textsf{IN},{|H|}$| if and only if |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime},1,|H|)} \in \textrm{ABD}_\leq\textsf{IN},{|H|}$| if and only if |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime},1,|H|)} \in \textrm{ABD}_=\textsf{IN},{|H|}$|⁠. The latter is true also when replacing |$|H|$| by |$|E|$| or |$|M|$|⁠. This proves the claimed |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hardnesses. (2.) From Fellows et al. [21, Prop. 4] we know that all three problems for |${\textsf{BR}}$| are |${\textbf{DP}} $|-complete for |$|H|=0$| even if |$|M|=1$|⁠. We argue that the hardness can be extended to |${\textsf{IN}}_2$|⁠. Note that |$\left \langle{\textsf{IN}}_2\cup \{F\} \right \rangle ={\textsf{BR}}$| where |$F(x) \equiv \neg x$|⁠. Creignou & Zanuttini [16] prove that |$\textrm{ABD}(S \cup \{F\}) \leq^{\textrm{P}}_m \textrm{ABD}(S \cup \{\texttt{SymOR}_{2,1}\})$| where |$\texttt{SymOR}_{2,1}(x,y,z)= ((x\rightarrow y) \land T(z)) \lor ((y\rightarrow x) \land F(z))$|⁠. Moreover, they also prove that |$\texttt{SymOR}_{2,1} \in \left \langle{S} \right \rangle $| such that |${\textsf{IN}}_2\subseteq \left \langle{S} \right \rangle $| [16, Lem. 21/27]. Finally, having |$|M|=1$| allows us to use their proof and, as a consequence, |$ \textrm{ABD}(\textrm{BR}) \leq ^{\textbf{P}}_m \textrm{ABD}(S)$| such that |${\textsf{IN}}_2 \subseteq \left \langle{S} \right \rangle $|⁠. This gives the desired lower bound for |${\textsf{IN}}_2$|⁠. Regarding |${\textsf{II}}_0$|⁠, the proof follows by a similar argument using the observations that |$\left \langle{\textsf{II}}_0 \cup{\{\, T \,\}} \right \rangle = {\textsf{BR}}$| and |$\texttt{OR}_{2,1} \in \left \langle{S} \right \rangle $| such that |${\textsf{II}}_0\subseteq \left \langle{S} \right \rangle $| where |$\texttt{OR}_{2,1}(x,y) = x\rightarrow y$| [16, Lem. 19/27]. (3.) Nordh and Zanuttini [31, Prop. 46/47] prove |${\boldsymbol{\varSigma _2^P}} $|-hardness for both |${\textsf{IN}}_2$| as well as |${\textsf{II}}_0$| with positive literal manifestations. This implies that the |$1$|-slice of each of |$\textrm{ABD}(\textsf{IN}_2,{|M|})$| and |$\textrm{ABD}(\textsf{II}_0,{|M|})$| is |${\boldsymbol{\varSigma _2^P}} $|-hard, which gives the desired result. For |$\textrm{ABD}_\leq(S, |M|)$| and |$\textrm{ABD}_=(S, |M|)$|⁠, the results follow from Lemma 3.7. Fixed-parameter tractable cases The following corollary is immediate because the classical questions corresponding to these cases are in |${\textbf{P}} $| due to Nordh and Zanuttini [31]. Corollary 3.10 The problem |$\textrm{ABD}(S, k)$| is |${\textbf{FPT}} $| for any parameterization |$k$| and |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}$| with |${\textsf{C}} \in{\{\, {\textsf{IV}}_2, {\textsf{ID}}_1, {\textsf{IE}}_1, {\textsf{IS}_{12}} \,\}} $|⁠. The next result is already due to Fellows et al. [21, Prop. 13]. Corollary 3.11 The problems |$\textrm{ABD}(S, |V|)$|⁠, |$\textrm{ABD}_\leq(S, |V|)$|⁠, |$\textrm{ABD}_=(S, |V|)$| are all |${\textbf{FPT}} $| for all Boolean constraint languages |${S}$|⁠. Now, we prove |${\textbf{P}} $|-membership for some cases of the classical problems and start with the essentially positive cases. The proof idea is to start with unit propagation. The positive clauses do not explain anything and one just only checks whether the elements of |$M$| appear either in |${\textit{KB}}$| or |$H$|⁠. Then, we need to adjust the size accordingly. Lemma 3.12 The classical problems |$\textrm{ABD}_=(S)$| and |$\textrm{ABD}_\leq(S)$| are in P for |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{02}}$|⁠. Proof. We prove the claim for |$\textrm{ABD}_=(S)$|⁠, whereas the result for |$\textrm{ABD}_\leq(S)$| is due to Lemma 3.8. Let |${(V,H,M,{\textit{KB}},s)}\in \textrm{ABD}_=(S)$| be an instance where |${S}\subseteq{\textsf{IS}_{02}}$|⁠. Denote by |$ H^{\prime},M^{\prime},{\textit{KB}}^{\prime}$| the result of applying the unit propagation on those literals |$y$| such that |$y \in \textrm{Lit}({\textit{KB}}) \backslash (H^{+} \cup M^{-})$|⁠. Recall that for a set |$Y$| of literals, |$Y^{+}$| (resp., |$Y^{-}$|⁠) denotes the set of positive (negative) literals formed upon |$Y$|⁠. In unit propagation, for a unit clause |$u$|⁠, any clause containing |$u$| can be deleted and delete in any clause |$\sim \!u$|⁠, where |$\sim \!u=x$| if |$u=\lnot x$| is a negative literal and |$\sim \!u=\lnot x$| if |$u=x$| is a positive literal. Note that literals |$y \in H^{+}\cup M^{-}$| (i.e. |$y\in H$| or |$y=\neg m$| with |$ m \in M$|⁠) are excluded from this rule as mentioned above. The reason for this choice is as follows. If |$\neg m \in{\textit{KB}}$| for some |$m\in M$| then removing |$m$| from |${\textit{KB}}\cup M$| transforms a ‘no solution’—to a ‘yes solution’—instance. Similarly, removing an |$h\in H$| from |${\textit{KB}}\cup H$| may decrease the solution size of the instance. Finally, the positive literal |$m\in M$| may or may not be processed. However, it is important to consider |$h\in H^{-}$| since this helps in invalidating the clauses of length |$\geq 2$|⁠. Let |$P$| and |$N$| be the positive, respectively negative unit clauses of |${\textit{KB}}^{\prime}$| over |$\left \langle{S} \right \rangle $|⁠. Note that if |$N\not =\emptyset $| then there can be no explanation for |$M$|⁠. This is due to the fact that only negative unprocessed literals are over |$M$| implying that |${\textit{KB}}$| is inconsistent with |$M$|⁠. Because of this, we have |$N=\emptyset $|⁠. Moreover, the positive clauses of length |$\geq 2$| in |${\textit{KB}}^{\prime}$| do not explain anything as a variable cannot be enforced |$0$|⁠. Therefore, a positive literal |$x$| cannot explain anything more than |$x$| itself. This implies that there is an explanation for |$M$| if and only if |$M^{\prime} \subseteq H^{\prime} \cup P$|⁠. Now, the set |$M^{\prime}\setminus P$| denotes those |$m\in M$| that are not already explained by |${\textit{KB}}$| and must be explained by |$H^{\prime}$|⁠. As a consequence, there exists an explanation for |$\textrm{ABD}_\leq(S)$| if and only if |$M^{\prime}\setminus P \subseteq H^{\prime}$| and |$|M^{\prime}\setminus P| \leq s$|⁠. The consistency is already assured by the fact that |$N=\emptyset $|⁠. Finally, to determine whether there is an explanation |$E \subseteq H$| of size |$s$|⁠, it suffices to check additionally whether |$|H^{\prime}| \geq s$|⁠. This argument ensures whether we can artificially increase the solution size, since, in that case an |$E \subseteq H^{\prime}$| with above conditions constitutes an explanation for the problem . If this is not true, then no explanation of size |$s$| exists. The unit propagation and the size comparisons can be done in polynomial time, which proves the claim. The following lemma proves that essentially negative languages for |$\textrm{ABD}_\leq$| also remain tractable. Lemma 3.13 The classical problem |$\textrm{ABD}_\leq(S)$| is in P if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠. Proof. First, we prove the result with respect to |$\left \langle S \right \rangle _{\neq }\subseteq{\textsf{IS}_{12}}$|⁠. Let |$P$| denote the set of positive unit clauses from |${\textit{KB}}$| and denote |$E_{MP} = M \setminus P$|⁠. Now, we have the following two observations. Observation 1 There exists an explanation iff |$E_{MP} \subseteq H$| and |$M$| is consistent with |${\textit{KB}}$|⁠. That is, what is not yet explained by |$P$| must be explainable directly by |$H$| because negative clauses cannot contribute to explaining anything, they can only contribute to ‘rule out’ certain subsets of |$H$| as possible explanations. Observation 2 If there exists an explanation, then any explanation contains |$E_{MP}$|⁠. As a result, |$E_{MP}$| represents a cardinality-minimal and a subset-minimal explanation. We conclude that there exists an explanation |$E$| with |$|E| \leq s$| iff |$E_{MP}$| constitutes an explanation and |$|E_{MP}| \leq s$|⁠. Now, we proceed with base independence for this case. Claim. |$\textrm{ABD}_\leq(S \cup \{=\}) \leq^{\textbf{P}}_m \textrm{ABD}_\leq(S)$| for |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠. Proof of Claim. The reduction gets rid of the equality clauses by removing them and deleting the duplicating occurrences of variables. This decreases only the size of |$H$| and might also the size of an explanation |$E$|⁠. Notice that |$x=y\in{\textit{KB}}$| does not enforce both |$x$| and |$y$| into |$E$|⁠. This completes the proof to the lemma. Finally, the |$2$|-affine cases are also tractable as we prove in the following lemma. The idea is, similar to Creignou et al. [10, Prop. 1], to change the representation of the knowledge base. Lemma 3.14 The classical problems |$\textrm{ABD}_=(S)$| and |$\textrm{ABD}_\leq(S)$| are in P if |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_1$|⁠. Proof. Analogously to Creignou et al. [10, Prop. 1], we change the representation of the |${\textit{KB}}$|⁠. Without loss of generality, suppose |${\textit{KB}}$| is satisfiable and contains no unit clauses since unit clauses can be dealt with in a straightforward way. Each clause expresses either equality or inequality between two variables. With the transitivity of the equality relation and the fact that (in the Boolean case) |$a \neq b \neq c$| implies |$a = c$|⁠, we can identify equivalence classes of variables such that each two classes are either independent or they must have contrary truth values. We call a pair of dependent equivalence classes |$(X, Y)$| a cluster (⁠|$X$| and |$Y$| must take contrary truth values). Denote by |$X_1, \dots , X_p$| the equivalence classes that contain variables from |$M$| such that |$X_i \cap M \neq \emptyset $|⁠. Denote by |$Y_1, \dots , Y_p$| the equivalence classes such that for each |$i$| the pair |$(X_i, Y_i)$| represents a cluster. We make the following stepwise observations. There is an explanation iff |$\forall i: H \cap X_i \neq \emptyset $|⁠. The size of a minimal explanation (⁠|$E_{min}$|⁠) is |$p$|⁠, it is constructed by taking exactly one representative from each |$X_i$|⁠. There exists an explanation of size |$\leq s$| iff |$p \leq s$|⁠. An explanation of maximal size (⁠|$E_{max}$|⁠) can be constructed as follows: (a) |$E:= \emptyset $|⁠, (b) for each |$i$| add to |$E$| all variables from |$X_i \cap H$|⁠, (c) for each cluster |$(X,Y) \notin \{(X_i, Y_i) \mid 1 \leq i \leq p\}$|⁠: if |$|X\cap H| \geq |Y\cap H|$|⁠: add to |$E$| the set |$X\cap H$|⁠, else: add to |$E$| the set |$Y\cap H$|⁠. Any explanation size between |$|E_{min}|$| and |$|E_{max}|$| can be constructed. There is an explanation of size |$=s$| iff |$|E_{min}| \leq s \leq |E_{max}|$|⁠. This completes the proof. Lemmas 3.12–3.14 imply the following corollary. Corollary 3.15 The following problems are FPT for any |$k\in{\{\, |H|,|E|, |M| \,\}} $|⁠. |$\textrm{ABD}_=(S, k)$| if |$\left \langle{S} \right \rangle \subseteq C$| for |${\textsf{C}} \in{\{\, {\textsf{IS}_{02}},{\textsf{ID}}_1 \,\}} $|⁠, |$\textrm{ABD}_\leq(S, k)$| if |$\left \langle{S} \right \rangle \subseteq C$| for |${\textsf{C}} \in{\{\, {\textsf{IS}_{02}},{\textsf{ID}}_1,{\textsf{IS}_{12}} \,\}} $|⁠. Now we move towards the parameter specific results for each problem. 3.3 Parameter ‘number of hypotheses’ |$|H|$| For this parameter, it turns out that the only intractable cases are those pointed out in Lemma 3.9. Theoem 3.16 |$\textrm{ABD}(S, |H|)$|⁠, |$\textrm{ABD}_\leq(S, |H|)$| and |$\textrm{ABD}_=(S, |H|)$| are |${\textbf{para-}}{\textbf{DP}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$|⁠, |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in{\{\, {\textsf{IE}}_2, {\textsf{IV}}_2, {\textsf{ID}}_2,{\textsf{IL}}_2 \,\}} $|⁠. Proof. (1.+2.) Follows from Lemma 3.9. (3.) Recall that |${\textrm{SAT}}({S})\ \textrm{and}\ {\textrm{IMP}}({S})$| are both in |${\textbf{P}} $| for every |${S}$| in the question (Prop. 2.7). By |$|H|\ge |E|$|⁠, we have that |$\binom{|H|}{|E|}=|H|^{|E|}\in O(k^k)$|⁠, where |$k=|H|$|⁠. Consequently, we brute-force the candidates for |$E$| and verify them in polynomial time. This yields |${\textbf{FPT}} $| membership. 3.4 Parameter ‘number of explanations’ |$|E|$| In this subsection, we consider the solution size as a parameter. Notice that, because of the parameter |$|E|$|⁠, the parameterized version of the problem |$\textrm{ABD}$| is not meaningful anymore. As a result, we only consider the size limited variants |$\textrm{ABD}_=$| and |$\textrm{ABD}_\leq $|⁠. The following theorem provides a classification of both problems into six different complexity degrees. Theoem 3.17 The problems |$\textrm{ABD}_\leq(S, |E|)$| and |$\textrm{ABD}_=(S, |E|)$| are |${\textbf{para-}}{\textbf{DP}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$| |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, |${\textbf{W}\textbf{P}}$|-complete if |${\textsf{IE}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IE}}_2$|⁠, |$\textbf{W}[2]$|-complete if |${\textsf{IM}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{C}}$| for |${\textsf{C}} \in \{{\textsf{ID}}_2,{\textsf{IS}^{\ell }_{10}}, {\textsf{IV}}_2\}$|⁠, |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_1$| or |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{02}}$|⁠, Moreover, if |${\textsf{IS}^{2}_{1}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}} $|⁠, then |$\textrm{ABD}_\leq(S, |E|) \in \textbf{FPT}$| and |$\textrm{ABD}_=(S, |E|)$| is |$\textbf{W}[1]$|-complete. Proof. (1.+2.) Follows from Lemma 3.9. (3.) The upper bound for |${\textsf{IE}}_2$| follows from the fact that |${\textrm{SAT}}({\textsf{IE}}_2)$| and |${\textrm{IMP}}({\textsf{IE}}_2)$| are in |${\textbf{P}} $| (cf. Prop. 2.7). Guessing |$E$| takes |$k \cdot \log n$| non-deterministic steps and verification can be done in polynomial time. For the lower bound, we argue that the proof from [21, Cor. 9] for definite Horn theories (⁠|${\textsf{IE}}_1$|⁠) can be extended. The only types of clauses used are |$x \land y \rightarrow z$| and |$x \rightarrow y$|⁠, which are both in |${\textsf{IE}}$| and consequently expressible by |${S}$| as |${\textsf{IE}} \subseteq \left \langle{S} \right \rangle $|⁠. Both membership and hardness arguments are valid for |$\textrm{ABD}_\leq(S, |E|)$| as well (the problem in [21, Cor. 9] used for hardness is Monotone Circuit SAT, which is monotone). (4.) The completeness for |$\textrm{ABD}_=(S, |E|)$| such that |$\left \langle{S} \right \rangle ={\textsf{IM}}$| follows from Lemma 3.18. Lemma 3.19 strengthens the result by showing -membership of the problem |$\textrm{ABD}_=(S, |E|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. The question |$\textrm{ABD}_\leq(S, |E|)$| for the above two cases follow from the monotone argument of Lemma 3.8. Moreover, for |${\textsf{ID}}_2$|⁠, the result follows from [21, Thm. 21]. The membership for |$\textrm{ABD}_=(S, |E|)$| and |$\textrm{ABD}_\leq(S, |E|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$| is due to Lemmas 3.20 and 3.21, respectively. The hardness for both cases follows from Lemma 3.18. (5.) Follows from Corollary 3.15. Finally, the |${\textbf{FPT}} $| membership for |$\textrm{ABD}_\leq\textsf{IS}_{12}, {|E|}$| is shown in Corollary 3.15. The |$\textbf{W}[1]$|-hardness for |$\textrm{ABD}_=(S, |E|)$| with |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$| follows from Lemma 3.22 where we prove |$\textbf{W}[1]$|-hardness for the languages |${S}$|⁠, such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠. The |$\textbf{W}[1]$| membership for |$\textrm{ABD}_=(S, |E|)$| with |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠, this means also the arbitrary bases, is due to Lemma 3.23 This concludes all the cases in Theorem 3.17. 3.4.1 Intermediate Lemmas Note that the difficult part of the abduction problem for |${S}$| such that |${\textsf{IM}} \subseteq \left \langle{S} \right \rangle $| is the case when a solution of size larger than |$k$| is found. This solution must be reduced to one of size |$\leq k$| (resp. |$=k$|⁠). First we prove the completeness for the language |${S}$| such that |$\left \langle{S} \right \rangle ={\textsf{IM}}$|⁠. Later we extend the membership result to the languages in |$ {\textsf{IV}}_2$| and |${\textsf{IS}^{l}_{10}}$|⁠. Lemma 3.18 |$\textrm{ABD}_=(S, |E|)$| is |$\textbf{W}[2]$|-complete if |$\left \langle{S} \right \rangle ={\textsf{IM}}$|⁠. Proof. For membership we prove that |$\textrm{ABD}_=(\textsf{IM}, |E|) \leq^{\textbf{FPT}} {\textrm{p-WSAT}(\varGamma _{2,1})}$|⁠. The latter is known to be |$\textbf{W}[2]$|-complete (Proposition 2.4). Let |${(V,H,M,{\textit{KB}},k)}$| be an instance of |$\textrm{ABD}_=(\textsf{IM}, |E|)$|⁠, where the solution size is the parameter (i.e. |$s=k$|⁠). Specifically, let |${\textit{KB}}=\bigwedge \limits _{i\leq r}(x_i \rightarrow y_i)$| and |$M= m_1\land \ldots \land m_{|M|}$|⁠. Note that, in order to explain a single |$m_i\in M$|⁠, a single |$h\in H$| suffices. As a result, for each |$m_i\in M$| we associate a set |$H_i \subseteq H$| of hypotheses that explains |$m_i$|⁠. This implies that every element (singleton subset) of |$H_i$| explains |$m_i$|⁠. Now, it is enough to check that at least one such |$h\in H_i$| can be selected for each |$m_i$|⁠. For this we map |${(V,H,M,{\textit{KB}},k)}$| to |$(\phi ,k)$| where |$\phi = \bigwedge \limits _{i\leq |M|} \bigvee \limits _{x\in H_i} x$|⁠. Then our claim is that |${(V,H,M,{\textit{KB}},k)}$| has an explanation |$E$| if and only if |$\phi $| has a satisfying assignment of size |$k$|⁠. Clearly, there is a |$1-1$|-correspondence between solutions |$E$| of |${(V,H,M,{\textit{KB}},k)}$| and satisfying assignments |$\theta $| with weight |$k$| for |$\phi $|⁠. That is, |$\theta (x)=1 \iff x\in E$|⁠. For hardness, we reduce from |${\textrm{p-WSAT}(\varGamma ^+_{2,1})}$|⁠, which is |$\textbf{W}[2]$|-complete by Proposition 2.4. Given |$\bigwedge \limits _{i\in q}\bigvee \limits _{j\in r}(X_{ij}^+) = \bigwedge \limits _{i\in q}(X_{i_1}\lor \ldots \lor X_{i_r})$|⁠, where |$\textrm{var}(\alpha )={\{\, X_{ij}\mid i\in q, j\in r \,\}} $|⁠, we let |${\textit{KB}}= \bigwedge \limits _{i\in q}\bigwedge \limits _{x\in h_i}(x\rightarrow h_i$|⁠, |$H=\textrm{var}(\alpha )$|⁠, |$M= \bigwedge \limits _{i\in q}h_i $| and |$V=H\cup M$|⁠. Then for a subset |$E\subseteq H$| we have that, |$E$| is an explanation for |$\textrm{ABD}_=(\textsf{IM}, |E|) \iff \theta \models \phi $| where |$\theta (x)=1 \iff x\in E$|⁠. Now we show that with a little modification, the same reduction (as for the membership of |${\textsf{IM}}$|⁠) can be used to prove |$\textbf{W}[2]$|-membership for |${\textsf{IV}}_2$|⁠. Lemma 3.19 |$\textrm{ABD}_=(S, |E|)$| is in |$\textbf{W}[2]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. Proof. In the IM-case, we dealt only with clauses of type |$x \rightarrow y$|⁠. We refer to such classes as type-|$0$| clauses. In |${\textsf{IV}}_2$| we have additional clauses of the following types: Unit clauses: both positive and negative. |$x$|⁠, |$\neg x$| Positive clauses of size two or greater: |$(x_1 \lor \dots \lor x_n)$|⁠, |$n\geq 2$| Clauses with exactly one negative literal of size |$3$| or greater: |$(\neg x_0 \lor x_1 \lor \dots \lor x_n)$|⁠, |$n\geq 2$| We can we eliminate the type-|$1$| clauses by unit propagation and obtain thereby a satisfiability equivalent formula. Note that this transformation process can generate additional clauses of type-|$0$|⁠, type-|$2$| or type-|$3$|⁠. As a consequence, we end up only with clauses of either type-|$0$|⁠, type-|$2$| or type-|$3$| and, particularly, no type-|$1$| clauses anymore. This transformation does not preserve all the satisfying assignments but those can be maintained by adding fixed values of the eliminated variables to the assignment. Now, we argue that by applying resolution on the variables in |${\textit{KB}} \setminus H$|⁠, we can ignore type-2 and type-3 clauses. Notice that we do not apply resolution to the variables in |$H$|⁠. Recall the idea behind the construction of Lemma 3.18, we want to come up with a formula that has a satisfying assignment if and only if our abduction instance has an explanation. A satisfying assignment that selects a variable |$x \in H$| (maps |$x$| to |$1$|⁠) forces all the variables |$y_1,\ldots , y_n$| such that |$(x\rightarrow y_i) \in{\textit{KB}}$| to be mapped |$1$| for |$i\leq n$|⁠. Furthermore, it also forces each |$z_{i,j}$| such that |$(y_i \rightarrow z_{i,j}) \in{\textit{KB}}$| to be mapped |$1$|⁠, and so on. This precisely captures the intuition that |$x$| (as a hypothesis) explains each |$y_i$| and |$z_{i,j}$|⁠. As a consequence, removing such variables from |$H$| (owing to resolution) in the case when those variables explain some manifestation would be problematic. Finally, we prove the claim that for |$\textrm{ABD}_=(\textsf{IV}_2, |E|)$| we can ignore the type-|$2$| and type-|$3$| clauses. Type-2 clauses are irrelevant since the satisfaction of such clauses does not force any particular variable to |$1$|⁠. In a type-3 clause (⁠|$C= \neg x_1 \lor x_2 \lor \dots \lor x_m$|⁠) the variable |$x_1$| forces a whole clause to be true (at least one of the remaining variables must be mapped to |$1$|⁠). Such clauses cannot be ignored right away because there might be further clauses of the form |$\neg x_j \lor m$| for each |$2\leq j \leq m $| with |$m\in M$| and an explanation to |$m$| might be lost (selecting |$x_1$| in the solution). However, after applying resolution we know that type-3 clauses only force one of the many positive variables to |$1$| and do not actually force a single variable to |$1$|⁠. As a result, this allows us to ignore type-3 clauses as well. Consequently, we are only left with type-0 clauses. This completes the proof by the same arguments as in the proof of Lemma 3.18. The question |$\textrm{ABD}_\leq(S, |E|)$| for the above two cases follows from the monotone argument of Lemma 3.8. Moving forward, the hardness for |${\textsf{IS}^{\ell }_{10}}$| is a consequence of the |$\textbf{W}[2]$|-hardness for IM. However, we strengthen this results for |$\textrm{ABD}_=$| to |$\textbf{W}[2]$|-completeness by showing the membership in |$\textbf{W}[2]$|⁠. Lemma 3.20 Let |$\ell \ge 2$|⁠, then |$\textrm{ABD}_=(S, |E|)$| is in |$\textbf{W}[2]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. We reduce our problem to |${\textrm{p-WSAT}(\varGamma _{2,1})}$|⁠, which is |$\textbf{W}[2]$|-complete (Prop. 2.4). Consider the reduction from Lemma 3.18 again, where we map |${(V,H,M,{\textit{KB}},s)}$| to |$(\phi ,k)$|⁠, where |$\phi = \bigwedge \limits _{i\leq m} \bigvee \limits _{x\in H_i} x$|⁠. The only difference from Lemma 3.18 is that in |${\textsf{IS}^{\ell }_{10}}$| there are additional constraints of the form |$(\neg x_1 \lor \ldots \lor \neg x_q)$| where |$q\leq \ell $|⁠. Now we have two cases. If all the additional constraints contain exclusively variables from H then we simply add these constraints to |$\phi $| and obtain a new formula |$\psi $|⁠. Since any satisfying assignment for |$\psi $| would satisfy these constraints as well as |$\phi $| and therefore is an explanation as required. Conversely, any explanation would yield a satisfying assignment for this new formula |$\psi $| since this explanation is consistent with |${\textit{KB}}$|⁠. Now suppose that constraints contain variables that are not from |$H$|⁠. We transform such constraints into their equivalents, which contain variables only from H. To achieve this we repeat the following procedure as long as applicable: Pick a variable |$u \not \in H$| occurring in a constraint |$C_u$|⁠. Compute the set of hypotheses |$H_u \subseteq H$| that explain |$u$| (analogously to Lemma 3.18). Let |$H_u = {\{\, h_1,\ldots , h_r \,\}} $|⁠. Now we replace the constraint |$C_u$| by |$r$| copies of itself and in each |$C_u^i$| we replace the variable |$u$| by |$h_i$|⁠. Note that this does not change the width of any clause. Finally, we add these clauses to |$\phi $| and obtain a new formula |$\psi $|⁠. Claim. The above construction preserves the correspondence between the solutions of |$\textrm{ABD}_=({\textsf{IS}^\ell_{10}}, |E|)$| and the satisfying assignments of |$\phi $| with weight |$k$|⁠. Moreover, it can be achieved in polynomial time. Proof of Claim. Note that the difference between Lemma 3.18 and this case is in the fact that a solution to |$\textrm{ABD}_=({\textsf{IS}^\ell_{10}}, |E|)$| must satisfy additional constraints as specified above. The problematic part is when some variables |$x_i,\ldots x_j$| are in |$H$| and some constraint over these variables appears in the |${\textit{KB}}$|⁠. The formula |$\psi $| must not allow such elements to be the part of solution since the constraints stop certain elements to appear together in a solution (being negative clauses). This proves the first claim in conjunction with the arguments in Lemma 3.18. Now we prove that this transformation works in polynomial time. The worst case is when a clause contains no variable from H. Furthermore assume that this clause is of maximum arity, say |$C= (\neg x_1,\ldots \neg x_q)$| where |$q\leq \ell $| and |$q$| is the maximum arity of constraint language in |${\textit{KB}}$|⁠. Each |$x_i$| can have the associated set |$H_{x_i}$| of maximum size |$n$| where |$n$| is input size. Hence each clause will be blown-up to at most |$n^q$| new constraints at the completion of the above procedure. As |$q$| is constant (only depends on the constraint language and not on the input), the factor |$n^q$| is polynomial. Since there are polynomial many constraints to check for this procedure, we conclude that the transformation takes only polynomial time. Eventually, similar arguments as in Lemma 3.18 for |$\psi $| complete the proof. We wish to point out that the reduction in Lemma 3.20 does not immediately settle the complexity for |$\textrm{ABD}_\leq(S, |E|)$| for |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. In the following lemma, we reduce |$\textrm{ABD}_\leq(S, |E|)$| to |${\textrm{Short-NTM-Halt}}$|⁠, the halting problem for non-deterministic multi-tape Turing machines. The input to the problem is |$(\mathbb M, k)$|⁠, where |$\mathbb M$| is a non-deterministic Turing machine, |$k$| is the parameter and the task is to decide whether |$\mathbb M$| accepts the empty string in at most |$k$| steps. This problem is |$\textbf{W}[2]$|-complete [22, Thm. 7.28]. The following reduction provides the |$\textbf{W}[2]$|-membership for |$\textrm{ABD}_\leq(S, |E|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠, the hardness follows from Lemma 3.18. Lemma 3.21 Let |$\ell \ge 2$|⁠, then |$\textrm{ABD}_\leq(S, |E|)$| is in |$\textbf{W}[2]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. The proof is in fact an extension of Lemma 3.20. Proceeding as before, we map |${(V,H,M,{\textit{KB}},k)}$| to |$(\psi ,k)$|⁠, where |$\psi $| is a collection of positive and negative clauses. Let |$U_i$| for |$i\leq P$| (resp., |$V_j, j\leq N$|⁠) denote the collection of positive (negative) clauses and |$L=P+N$|⁠. In the following, we reduce our problem to the problem |${\textrm{Short-NTM-Halt}}$| for multi-tape Turing machines. The reduction provides a multi-tape NTM |$\mathbb{M}$| such that |$\psi $| has a satisfying assignment of size at most |$k$| if and only if |$\mathbb{M}$| accepts the empty string in at most |$f(k)$| steps. |$\mathbb{M}$| has |$L+1$| tapes and there is one tape per each clause. The initial |$P$| tapes are dedicated to the positive clauses and the following |$N$| tapes to the negative ones. For convenience, we name the |$i$|th tape corresponding to a positive clause, as |$u_i$| (⁠|$i\leq P$|⁠) and |$j$|th tape corresponding to a negative clause, as |$v_j$| (⁠|$j\leq N$|⁠). The last tape is referred to as the tape |$L+1$|⁠. Intuitively, the computation of |$\mathbb{M}$| has the following four phases. Mark the length of each negative clause |$V_j$| on the tape |$v_j$|⁠, where |$j\leq N$|⁠. At the same time, write non-deterministically |$k$| elements |$x_1,\ldots ,x_k$| from |$V$| on the tape |$L+1$|⁠. Remove the duplicate variables from the tape |$L+1$|⁠. Read the tape |$L+1$|⁠. At the same time, for each tape |$w_r$|⁠, mark the cells if the elements being read appears in the respective clause |$W_r$| where |$r\leq L$|⁠. If at least one cell of each tape |$u_i$| (⁠|$i\leq P$|⁠) is marked and not every cell of the tape |$v_j$| (⁠|$j\leq N$|⁠) is marked, then accept. We first claim that the negative clauses of length |$\geq k$| can be ignored. The reason is that, for any assignment that maps |$k$| variables to |$1$|⁠, there are still literals in such clauses that are mapped to |$0$|⁠. Consequently, these clauses are trivially satisfied. This implies that the length of each tape corresponding to a negative clause is bounded by |$k$|⁠. For positive clauses, the length does not matter. This is because, our machine only needs to determine if at least one variable appearing in each positive clause is mapped to |$1$| in the assignment. Consequently, the machine only reads at most |$k$| cells on each of its tape. Our construction requires that the length of each |$V_j$| is hardcoded on the tape |$v_j$| for |$j\leq N$|⁠. This ensures that |$\mathbb{M}$| runs in parallel and does not need a state set of exponential size to ensure the correct computation. For each |$V_j$| of length |$l_j$|⁠, the length of the tape |$v_j$| is |$l_j+1$|⁠, where |$j\leq N$|⁠. This is achieved through having a collection of |$r+1$| states, where |$r = \textrm{max}\left |l_j \mid j \leq N\right |$| and |$r\leq k$|⁠. Moreover, even though |$\mathbb M$| can guess duplicate elements, it must work with the distinct collection of variables in the subsequent steps. That is, the multiple occurrences of any variable should be removed from the guessed assignment. A detailed but high-level description of |$\mathbb M$| is given below. – In the first |$k$| steps, the head of each tape |$v_j $| writes the symbol ‘*’ for |$l_j$|-cells and the symbol ‘#’ in the cell |$l_j+1$|⁠, where |$l_j$| is the length of |$V_j$| for |$j\leq N$|⁠. In the same steps, the head of the tape |$(L + 1)$| non-deterministically writes |$k$| elements |$x_1, \ldots , x_k$| from |$V$|⁠, into the first |$k$| cells. After |$k$| steps, the heads go back to the first cell. In the next (at most) |$k^2$| steps, the head on the tape |$L+1$| removes any duplicates. In the following |$h$| steps (where |$h\leq k$|⁠), the head of the tape |$(L + 1)$| reads the guessed elements (without duplicates) |$x_1, \ldots , x_h$| and at the same time, in |$r$|th of these steps, the head of each tape |$u_i$| determines whether |$U_i$| contains |$x_r$|⁠, for |$i\leq P $|⁠, while the head of each tape |$v_j$| determines if |$\neg x_r$| appears in |$V_j$| for |$j\leq N $|⁠. In the first case, it writes ‘|$yes$|’ in the cell, in the latter cases, it does not write. After writing in each case, the head moves right. If the variables do not appear in the corresponding clauses, the heads neither move nor print. After the previous |$h\leq k$| steps, the head on tapes |$u_i$| for |$i\leq P$| moves one step left while for other tapes, it stays in the same cell. If the head reads ‘|$yes$|’ in the |$u_i$|-tapes (⁠|$i\leq P$|⁠) and a ‘*’ in |$v_j$|-tapes (⁠|$j\leq N$|⁠), then |$\mathbb M$| accepts. Claim. |$\mathbb{M}$| can be constructed from |$\psi $| in FPT-time. Moreover, |$\mathbb M$| accepts the empty input in at most |$ k^2 + 3k+2$| steps if and only if |$\psi $| has a satisfying assignment of size at most |$k$|⁠. Proof of Claim. The number of tapes is |$L+1$| where |$L$| is the number of clauses in |$\psi $|⁠. The alphabet of |$\mathbb M$| constitutes |$V\cup \{yes,*,\#\}$| where |$V$| is the collection of variables in |$\psi $|⁠. The set of states has size |$k+O(1)$|⁠, |$k$| of which are required to ensure the first phase of the computation, i.e. guessing |$k$| elements and marking the tapes |$v_j$| for |$j\leq N$|⁠. Recall that the computation of |$\mathbb M$| completes in four phases. In the first phase, the head on the tape |$L+1$| moves right, writing |$x\in V$| (non-deterministically). Whereas, the head on the tape |$v_j$| (⁠|$j\leq N$|⁠) writes a ‘*’ for |$l_j$| many cells where |$l_j\leq k$| is the length of |$V_j$| and a ‘#’ in the last cell; moreover, the head stays in the last cell. When moving backwards, the head on the tape |$L+1$| can read any element |$x\in V$|⁠. However, the head on each |$v_j$|-tape (⁠|$j\leq N$|⁠) reads the symbol ‘#’ exactly once and the symbol ‘*’ in the remaining cells. The transitions for |$\mathbb M$| force every head to move one step left when reading ‘#’ and then the heads can only read the symbol ‘*’. This implies that the transition relation has the size |$O(k\cdot |\psi |^2)$| for the first phase. Finally, the transitions for ‘removing duplicates’, ‘comparing the variables’ and ‘the final check’ each has size |$O(k\cdot |\psi |)$|⁠. Therefore |$\mathbb M$| can be constructed from |$\psi $| in FPT-time. This proves the first part of the claim. For the second part, notice that the machine runs for |$2k + k^2 + k+2$| many steps. The first |$2k$| steps account for marking the length of each negative clause on the corresponding tapes and for guessing |$k$| elements on tape |$L+1$|⁠. In both cases the head of each tape should move back to read the first cell (the reason for |$2k$| steps). The following |$k^2$| steps are required to determine and remove the duplicate variables from the guessed list of variables. Lastly, at most |$k$| steps are required to compare the variables against each clause and the final two steps determine the accepting criteria for each tape. For the correctness, notice that the following three statements are equivalent. |$\mathbb M$| guesses |$k$| elements in such a way that the head of the tape |$u_i$| (⁠|$i\leq P$|⁠) reads ‘|$yes$|’, no head of the tape |$v_j$| (⁠|$j\leq N$|⁠) reads ‘#’ and the machine halts in the accepting state. The assignment (of weight at most |$k$|⁠) guessed by |$\mathbb M$| is such that there is at least one variable per each positive clause and for each negative clause, the assignment does not contain all of its variables. |$\psi $| has a satisfying assignment |$s$| of weight at most |$k$|⁠, |$s$| contains at least one variable from each positive clause and none of the negative clause contains all the variables appearing in |$s$|⁠. Consequently, if |$\psi $| has a satisfying assignment |$s$| of weight |$k$|⁠, then |$\mathbb M$| simply guesses this assignment and halts in the accepting state. Conversely, if |$\mathbb M$| accepts, then the guessed elements constitute a satisfying assignment for |$\psi $|⁠. This completes the proof to Lemma 3.21. Regarding the parameter |$|E|$|⁠, the only cases where |$\textrm{ABD}_\leq(S, |E|)$| and |$\textrm{ABD}_=(S, |E|)$| have different complexity is when |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠. The problem |$\textrm{ABD}_\leq(S, |E|)$| is FPT (Corollary 3.15). In the following lemmas, we prove |$\textbf{W}[1]$|-completeness for |$\textrm{ABD}_=(S, |E|)$|⁠. The |$\textbf{W}[1]$|-hardness for |$\textrm{ABD}_=(S, |E|)$| is proven for the languages |${S}$| such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠. Lemma 3.22 For any constraint language |${S}$| such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠, the problem |$\textrm{ABD}_=(S, |E|)$| is |$\textbf{W}[1]$|-hard. Proof. The problem IndependentSet is known to be |$\textbf{W}[1]$|-hard [17]. We reduce IndependentSet to |$\textrm{ABD}_=(S, |E|)$|⁠. Let |$((V, \tilde E), k)$| be an instance of |$p$|-IndependentSet and |$k$| the parameter. We map it to |${(V,H,M,{\textit{KB}},k+1)}$|⁠, where $$\begin{align*} {\textit{KB}} &:= \{(\neg x \lor \neg y) \mid (x,y) \in \tilde E\},\\ H &:= \textrm{var}({\textit{KB}}) \cup \{z\},\\ M &:= z. \end{align*}$$ Let |$U$| be an independent set of size |$k$| then |$U\land{\textit{KB}}$| is consistent because no two elements with an edge are in |$U$|⁠. As a consequence, |$U\cup \{z\}$| is an explanation for |${(V,H,M,{\textit{KB}},k+1)}$|⁠. Conversely, an explanation |$E$| for |${(V,H,M,{\textit{KB}},k+1)}$| of size |$k+1$| must include |$z$| as well as |$k$| other variables. Now, |$E \land{\textit{KB}}$| is consistent and this implies that no variables in |$E$| have an edge, consequently giving an independent set of size |$k$|⁠. This implies that |$(V, \tilde E)$| admits an independent set of size |$k$| if and only if |${(V,H,M,{\textit{KB}})}$| admits an explanation of size |$k+1$|⁠. Now we prove |$\textbf{W}[1]$| membership for |$\textrm{ABD}_=(S, |E|)$| with |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{12}}$|⁠, this means also arbitrary bases, in the lemma below. Lemma 3.23 Let |$\ell \ge 2$|⁠, then |$\textrm{ABD}_=(S, |E|)$| is in |$\textbf{W}[1]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{12}}$|⁠. Proof. We reduce |$\textrm{ABD}_=(S, |E|)$| to |${\textrm{p-WSAT}(\varGamma _{1,\ell })}$|⁠, which is |$\textbf{W}[1]$|-complete due to Prop. 2.4. Note that |$\varGamma _{1,\ell }$| is the class of |$\ell $|-CNF formulas. We want to mention here that the proof is correct even in the presence of equality constraints. As a consequence, the base independence is not implied by any of the previous lemmas but it follows due the proof below. According to Lemma 3.13, we can determine whether there exists a solution of size |$\leq s$| in polynomial time. Let |${(V,H,M,{\textit{KB}},k)}$| be an instance of |$\textrm{ABD}_=({\textsf{IS}^\ell_{12}}, |E|)$| with |${\textit{KB}} = \bigwedge _{i\leq r} C_i \land N \land P \land E$|⁠, where |$C_i = (\neg{x^i_1} \lor \dots \lor \neg{x^i_\ell })$|⁠, and |$P, N$| denote the positive and negative unit clauses, respectively, and |$E$| are the equality clauses. Without loss of generality, assume that |${(V,H,M,{\textit{KB}},k)}$| admits a solution of size |$\leq k$| (otherwise, map it to a negative dummy instance). Moreover, it follows from Lemma 3.13 that in this case any solution |$E$| satisfies that |$E_{MP} \subseteq E \subseteq H$|⁠. This implies |$s \geq |E_{MP}|$|⁠. We also know from Lemma 3.13 that |$E_{MP}$| is an explanation for |$M$| and that both |$E_{MP}$| and |$M$| are consistent with all clauses in |${\textit{KB}}$|⁠. The question now reduces to whether we can extend |$E_{MP}$| to a solution of size |$k$| by adding |$k - |E_{MP}|$| variables from |$H \setminus E_{MP}$|? We show that this can be achieved and map |${(V,H,M,{\textit{KB}},k)}$| to |$\langle \varphi , k - |E_{MP}|\rangle $|⁠, where |$\varphi $| is obtained from |${\textit{KB}}$| by the following consecutive steps: first we take care of equality clauses. For each |$x_i=x_j \in{\textit{KB}}$|⁠, such that |$x_i, x_j \in H$|⁠, add to |$\phi $| the clauses |$(\neg x_i\lor x_j)$| and |$( x_i\lor \neg x_j)$|⁠. We add these two clauses to |$\phi $| ensuring that corresponding to each clause of the form |$x_i=x_j$|⁠, either both |$x_i, x_j$| are in the solution, or none is. 1. Remove all clauses |$C_i$| containing only variables not from |$H$|⁠. Remove all negative unit clauses |$(\neg x)\in N$| such that |$x \notin H$|⁠. Note that after this step all remaining negative unit clauses are built upon variables from |$H \setminus E_{MP}$| only. For each clause |$C_i$|⁠, denote by |$X^i_H$| (resp., |$X^i_{\overline{H}}$|⁠) the variables from |$H$| (resp., not from |$H$|⁠). Execute the following: (a) Remove |$C_i$|⁠. (b) If |$X^i_{\overline{H}} \subseteq P$|⁠: add to |$\varphi $| the clause |$(\neg x_1 \lor \dots \lor \neg x_p)$|⁠, where |$\{x_1, \dots , x_p\} = X^i_H \setminus E_{MP}$|⁠. Otherwise nothing needs to be done as |$X^i_{\overline{H}}\not \subseteq P$| is true. Then, for some variable |$x\notin P$| we have that |$\lnot x\lor \bigvee _{x_j\in X^i_H}\lnot x_j$| is satisfiable via setting |$x$| to |$0$| if all |$x_j$| are mapped to |$1$|⁠. Note that after this step all remaining clauses |$C_i$| are built upon variables from |$H$| only. Remove all positive unit clauses |$(x) \in P$| such that |$x \notin H \setminus E_{MP}$|⁠. Note that after this step it holds that |$\textrm{var}(\varphi ) = H$| and all remaining positive unit clauses are built upon variables from |$H \setminus E_{MP}$| only. For all clauses |$C_i$|⁠: remove from |$C_i$| all literals built upon variables from |$E_{MP}$|⁠. Note that in the so obtained |$C_i^{\prime}$| at least one literal remains, because otherwise |$E_{MP}$| would be inconsistent with |$C_i$|⁠. After the last step has been implemented, it holds that |$\textrm{var}(\varphi ) = H \setminus E_{MP}$|⁠. As a consequence, the following equivalences are true: – |${(V,H,M,{\textit{KB}},k)} \textrm{ admits a solution of size exactly} k$|⁠. |$E_{MP}$| extends to a solution of size |$k$| by adding |$k - |E_{MP}|$| variables. |$\varphi $| has a satisfying assignment of size exactly |$k - |E_{MP}|$|⁠. This completes the proof to the lemma. 3.5 Parameter ‘number of manifestations’ |$|M|$| The complexity landscape regarding the parameter |$|M|$| is more diverse. The classification differs for each of the investigated problem variants. Consequently, we treat each case separately and start with the general abduction problem that provides a pentachotomy. Theoem 3.24 The problem |$\textrm{ABD}(S, |M|)$| is 1. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, 2. |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, 3. |${\textbf{para-}}{\textbf{NP}} $|-complete if |${\textsf{IE}}_0\subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IE}}_2 $|⁠, 4. |$\textbf{W}[1]$|-complete if |${\textsf{IS}^{2}_{11}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{C}}$| and |${\textsf{C}} \in \{{\textsf{ID}}_2, {\textsf{IS}^{\ell }_{10}}\}$|⁠, 5. |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in \{{\textsf{ID}}_1, {\textsf{IS}_{12}},{\textsf{IE}}_1,{\textsf{IV}}_2\}$|⁠. Proof. (1.+2.) We proved in Lemma 3.9 using the fact that 1-slice of each problem is hard for respective classes. (3.) The membership is trivial since the classical problem is NP-complete. For hardness, we prove that |$1$|-slice of the problem is NP-hard. Notice that due to Nordh and Zanuttini [31, Lemma 29] an abduction instance can be reduced to an instance with only one manifestation if the KB allows certain clauses. The idea is to encode a set of manifestations |$M$| by a single new manifestation |$y$|⁠, while adding the clause |$y\lor \bigvee \limits _{m\in M} \neg m$| to the |${\textit{KB}}$|⁠. Recall that |$M$| is a (positive) set of propositions, implying that the clause |$y\lor \bigvee \limits _{m\in M} \neg m$| is a Horn clause. Consequently, the aforementioned reduction is valid for |${\textit{KB}} \in{\textsf{IE}}_0$|⁠. This reduction in conjunction with the result for single literal manifestation ([31, Prop. 53]) implies that |$1$|-slice of the problem |$\textrm{ABD}(\textsf{IE}_0, |M|)$| is |${\textbf{NP}} $|-hard. Consequently, the problem is |${\textbf{para-}}{\textbf{NP}} $|-complete. (4.) The membership for |${S}$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_2$| follows from [21, Thm. 25]. Notice that the authors in [21] prove the completeness for the languages in |${\textsf{ID}}_2$| alone, but using the fact that the formulas (or clauses) in their reduction are |${\textsf{IS}^{2}_{11}}$|-formulas, we derive the hardness for |${\textsf{IS}^{2}_{11}}$|⁠. The |$\textbf{W}[1]$|-membership for the languages |${S}$| such that |$ \left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$| follows from Lemma 3.27. As a consequence, we have the desired completeness results. (5.) Follows from Corollary 3.10. For |$\textrm{ABD}_\leq$|⁠, definite Horn cases surprisingly behave different and are much harder than for the general case. Theoem 3.25 The problem |$\textrm{ABD}_leq(S, |M|)$| is 1. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, 2. |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, 3. |${\textbf{para-}}{\textbf{NP}} $|-complete if |${\textsf{IE}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IE}}_2$|⁠, 4. |$\textbf{W}[1]$|-complete if |${\textsf{IS}^{2}_{11}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{C}}$| and |${\textsf{C}} \in \{{\textsf{ID}}_2, {\textsf{IS}^{\ell }_{10}}\}$|⁠, 5. |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}\in \{{\textsf{ID}}_1, {\textsf{IS}_{12}}, {\textsf{IV}}_2\}$|⁠. Proof. (1.+2.) Follows from Theorem 3.24 in conjunction with Lemma 3.7. (3.) The membership is trivial since classical problem is in NP. For hardness, we reduce VertexCover to our problem similar to the approach of Fellows et al. [21, Thm. 5]. The problem can be translated into an abduction instance with |${\textit{KB}}\in{\textsf{IE}}$|⁠, consequently giving the desired hardness result. (4.) The membership for |${S}$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_2$| follows from [21, Thm. 25]. Notice that the authors in [21] prove the completeness for the languages in |${\textsf{ID}}_2$| alone, but using the fact that the formulas (or clauses) in their reduction are |${\textsf{IS}^{2}_{11}}$|-formulas, we derive the hardness for |${\textsf{IS}^{2}_{11}}$|⁠. The |$\textbf{W}[1]$|-membership for the languages |${S}$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$| follows from Lemma 3.28. (5.)FPT-membership for |$\textrm{ABD}_leq(S, |M|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IM}}$| follows from Lemma 3.29. Lemma 3.30 extends this result to |$\textrm{ABD}_leq(S, |M|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. The remaining cases are due to Corollary 3.15. Now, we end by stating the results for |$\textrm{ABD}_=$|⁠. Interesting to observe, the majority of the intractable cases is already much harder with large parts being |${\textbf{para-}}{\textbf{NP}} $|-complete. Even the case of the essentially negative co-clones that are FPT for |$\textrm{ABD}_\leq $| yield |${\textbf{para-}}{\textbf{NP}} $|-completeness in this situation. Merely the |$2$|-affine and dualHorn cases are |${\textbf{FPT}} $|⁠. Theoem 3.26 The problem |$\textrm{ABD}_=(S, |M|)$| is 1. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, 2. |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, 3. |${\textbf{para-}}{\textbf{NP}} $|-complete if |${\textsf{C}}_1 \subseteq \left \langle{S} \right \rangle $| where |${\textsf{C}}_1\in \{{\textsf{IS}^{2}_{1}},{\textsf{IE}}\}$| and |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}_2 \in \{{\textsf{IE}}_2, {\textsf{ID}}_2\}$|⁠, 4. |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in \{{\textsf{ID}}_1, {\textsf{IV}}_2\}$|⁠. Proof. (1.+2.) Follow from Theorem 3.24 in conjunction with Lemma 3.7. (3.) The membership in each case is trivial since the classical problems are NP-complete. The hardness for |${\textsf{IE}}\subseteq \left \langle{S} \right \rangle $| follows from the argument used in the proof of Theorem 3.25 for the |${\textsf{IE}}$| case. The hardness for the remaining cases follows from Lemma 3.31 where we prove that the problem |$\textrm{ABD}_=(S, |M|)$| is |${\textbf{para-}}{\textbf{NP}} $|-hard as long as |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠. The case for |$\textrm{ABD}_=({\textsf{IS}^{2}_{1}}, |M|)$|⁠, so also arbitrary bases, then follows as a corollary. (4.) The proof for |${\textsf{IV}}_2$| is due to the monotone argument of Lemma 3.8 and Theorem 3.25. The result for |${\textsf{ID}}_1$| is due to Corollary 3.15. 3.5.1 Intermediate lemmas Lemma 3.27 Let |$\ell \geq 2$| then the problem |$\textrm{ABD}(S, |M|)$| is in |$\textbf{W}[1]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. We use the same reduction as in Lemma 3.21 and argue that the so-obtained multi-tape Turing machine |$\mathbb M$| can be simulated by a single-tape machine |$\mathbb M^{\prime}$|⁠. This gives a reduction to the |$\textbf{W}[1]$|-complete problem |${\textrm{Short-NSTM-Halt}}$| [22, Thm. 6.17]. Moreover, the blow-up in the size of |$\mathbb M^{\prime}$| is only a function in the parameter |$k$| and |$\mathbb M^{\prime}$| runs for a number of steps that is bounded by a function in |$k$|⁠. In |$\textrm{ABD}$|⁠, there is no restriction on the solution size. As a consequence, the abduction instance has a solution if and only if the formula |$\psi $| of the reduction in Lemma 3.21 is satisfiable. Recall that the number of positive clauses in |$\psi $| is |$|M|$|⁠, which is the parameter |$k$| in this case. We claim that it is enough to determine if |$\psi $| has a satisfying assignment of weight at most |$k$|⁠. Claim. Let |$\phi $| be a |$\varGamma _{2,1}$|-formula with |$k$| positive and |$n$| negative clauses. Then |$\phi $| is satisfiable if and only if |$\phi $| has a satisfying assignment of size at most |$k$|⁠. Proof of Claim. The direction from right to left is trivial. For the other direction, note that if |$s\models \phi $| for some assignment |$s$| then, |$s \cap U_i \not = \emptyset $| for any |$i\leq k$| and |$ V_j \not \subseteq s $| for |$j\leq N$|⁠. Where we consider |$s$| as the collection of variables mapped to |$1$|⁠. Let |$s^{\prime}$| be the assignment obtained from |$s$| such that, for each positive clause |$U_i$|⁠, |$s^{\prime}$| selects exactly one variable from |$U_i$| (with repetition allowed for different clauses). Then |$s^{\prime}\models \phi $| and |$|s^{\prime}|\leq k$|⁠. This is because |$s^{\prime}$| selects exactly one variable from each positive clause and |$s^{\prime}\subseteq s$|⁠. As before, we ignore the negative clauses of length more than |$k$|⁠. This implies that there can be at most |$2^k$| negative clauses. Consequently, |$\mathbb M$| has at most |$k+2^k+1$| tapes. We argue that the size of |$\mathbb M^{\prime}$| is |$O(2^k\cdot k\cdot p(|\psi |))$| where |$p$| is some polynomial. This is because there are |$2^k+k+1$| tapes (the worst case) and consequently, the size of each transition is bounded by |$O(k+2^k)$|⁠. This implies that the size of |$\mathbb M^{\prime}$| is |$O((k+2^k)\cdot |\mathbb M|)$| where |$|\mathbb M|= O(k \cdot |\psi |^2)$|⁠. Moreover, |$\mathbb M^{\prime}$| runs for |$2^k\cdot f(k)^2$| steps where |$f(k)$| is the number of steps taken by |$\mathbb M$| (for details of the simulation see the textbook of Sipser [43, Theorem 7.8]). The correctness follows from Lemma 3.21. This completes the proof of the lemma. Lemma 3.28 Let |$\ell \geq 2$| then the problem |$\textrm{ABD}_\leq(S, |M|)$| is in |$\textbf{W}[1]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. Given |${(V,H,M,{\textit{KB}},s,|M|)}$|⁠, the task is to find an explanation of size at most |$s$| where |$|M|=k$| is the parameter. We argue that the reduction in Lemma 3.27 can be extended to this case. In Lemma 3.27 we proved that there is an explanation for |$\textrm{ABD}(S, |M|)$| if and only if there is an explanation of size |$|M|$| at most, where |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Now we have two cases. If |$k \leq s$| then the result holds already due to Lemma 3.27 (in particular the claim). This is because a solution of size |$k$| is also a solution of size at most |$s$| and there can be no solution of size in between |$k$| and |$s$| if there is no solution of size at most |$k$|⁠. On the other hand, if |$s< k$|⁠, the solution size is still bounded by the parameter. Our reduction in the proof of Lemma 3.27 takes care of this change by producing two parameters |$k_1$| and |$k_2$| such that |$k_1=k=|M|$| and |$k_2=s$|⁠. This refined reduction is still an FPT-reduction because the parameter |$f(k)=k_1+k_2$| is bounded by |$k$|⁠, i.e. |$f(k)\leq 2k$|⁠. The rest of the reduction remains the same. The only difference now is that the machine (say |$\mathbb M^{\prime\prime}$|⁠) guesses |$k_2$| elements and not |$k_1$| as |$\mathbb M^{\prime}$| does in Lemma 3.27. This completes the proof to the current lemma. We prove FPT-membership for |$\textrm{ABD}_\leq(S, |M|)$|⁠, where |$\left \langle{S} \right \rangle \subseteq{\textsf{IM}}$|⁠, by reducing our problem to the |${\textrm{MaxSATs}}$| problem that asks, given |$m$| clauses, is it possible to set at most |$s$| variables to true so that at least |$k$| clauses are satisfied. Lemma 3.29 The problem |$\textrm{ABD}_\leq(S, |M|)$| is FPT if |$\left \langle{S} \right \rangle \subseteq{\textsf{IM}}$|⁠. Proof. Given an instance |${(V,H,M,{\textit{KB}},s,|M|)}$| with |${\textit{KB}}=\bigwedge \limits _{i\leq r}(x_i \rightarrow y_i)$| and |$M= m_1\land \ldots \land m_{|M|}$|⁠. Recall that each |$m_i\in M$| can be explained by a single |$h_i\in H $|⁠. If |$|M|\leq s$| then there is nothing to prove. This is due the fact that in the proof of Lemma 3.18, there are fewer than |$s$| many sets of the form |$H_i$| each explaining an |$m_i\in M$|⁠. As a consequence, we need only select one |$h_{i,j}$| from each |$H_i$| as the part of a solution to yield a solution of size |$\leq s$|⁠. Accordingly, assume that |$|M|> s$|⁠. Proceed as in the proof of Lemma 3.18 and associate a set |$H_i \subseteq H$| of hypotheses with each |$m_i$| that explains it for |$i\leq |M|$|⁠. It is enough to check whether selecting at most |$s$| many elements |$h_i\in H$| can explain all the manifestations |$m_i \in M$|⁠. We reduce our problem to |${\textrm{MaxSATs}}$| [5] (we alter the notation slightly) asking, given a |$\textrm{CNF}$| formula on |$n$| variables with |$m$| clauses, if setting at most |$s$| variables to true satisfies at least |$k$| clauses. Let |$H^{\prime}$| be the collection of all |$H_i$|’s. For each |$i$| let |$C_i$| be the clause |$\bigvee \limits _{j} h_j$| where |$h_j\in H_i$|⁠. Furthermore, let |$C$| be the collection of all such clauses. Then |$C$| is built over variables in |$V^{\prime} = \bigcup _i H_i$|⁠. Our reduction maps |${(V,H,M,{\textit{KB}},s,|M|)}$| to |$\langle C, s, k \rangle $|⁠. Note that we only have |$|M|=k$| many clauses in |$C$| and, as a result, the question reduces to whether it is possible to set at most |$s$| variables from |$V^{\prime}$| to satisfy every clause in |$C$|? The reduced problem |${\textrm{MaxSATs}}$| when parameterized by |$k$| (the minimum number of clauses to be satisfied) is |${\textbf{FPT}} $| [5, Prop. 4.3]. Now, each |$H_i$| can be computed in polynomial time, the whole computation is a polynomial time reduction. Finally, the new parameter value |$k$| is exactly the same as the old parameter |$|M|$|⁠, the reduction is an |$\leq ^\textbf{FPT}$|-reduction. As a consequence, the lemma applies. We extend the |${\textbf{FPT}} $|-membership to languages in |${\textsf{IV}}_2$| using the same argument as in Lemma 3.19. Corollary 3.30 The problem |$\textrm{ABD}_\leq(S, |M|)$| is FPT if |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. Proof. After applying unit propagation and resolution we can ignore the positive clauses of length |$\geq 2$| and clauses with one negative literal of length |$\geq 3$|⁠. Lemma 3.31 For any constraint language |${S}$| such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠, the problem |$\textrm{ABD}_=(S, |E|)$| is |${\textbf{para-}}{\textbf{NP}} $|-hard. Proof. We prove that the |$1$|-slice of the problem is |${\textbf{NP}} $|-hard by reducing from classical |${\textrm{IndependentSet}} $| (which is |${\textbf{NP}} $|-complete [25]) to |$\textrm{ABD}_=(S)$|⁠. The reduction is essentially the classical counterpart of the one presented in Lemma 3.22. Let |$\langle V, \tilde E\rangle $| be an instance of |${\textrm{IndependentSet}} $|⁠. We map it to |${(V,H,M,{\textit{KB}},s)}$|⁠, where $$\begin{align*} {\textit{KB}} &:= \{(\neg x \lor \neg y) \mid (x,y) \in \tilde E\},\\ H &:= \textrm{var}({\textit{KB}}) \cup \{z\},\\ M &:= z,\\ s &:= k+1. \end{align*}$$ Then |$(V, \tilde E)$| admits an independent set of size |$k$| if and only if |${(V,H,M,{\textit{KB}},s)}$| admits an explanation of size |$s$|⁠. Notice that we did not mention the base independence for essentially negative languages in the previous proof. This is because, the |${\textbf{para-}}{\textbf{NP}} $|-membership as well as the base independence holds for |$\textrm{ABD}_=(S, |M|)$|⁠, where |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in \{{\textsf{IE}}_2,{\textsf{ID}}_2\}$|⁠. This gives the desired results for essentially negative languages as well. 4 Conclusion In this paper, we presented a two-dimensional classification of three central abductive reasoning problems (unrestricted explanation size, = and |$\leq $|⁠). In one dimension, we consider the different parameterizations |$|H|,|M|,|V|,|E|$|⁠, and in the other dimension we consider all possible constraint languages defined by corresponding co-clones except the affine co-clones. Often in the past, problems regarding the affine co-clones (resp., clones) resisted a complete classification [1, 2, 11, 13, 23, 27, 38, 44]. Also the result of Durand and Hermann [19] underlines how restive problems around affine functions are. It is difficult to explain why exactly these cases are so problematic but the notion of the Fourier expansion [32] of Boolean functions gives a nice and fitting view on that. Informally, the Fourier expansion of a Boolean function is a probability measure mimicking how likely a flip of a variable changes the function value. For instance, disjunctions have a very low Fourier expansion value, whereas the exclusive-or function has the maximum. Affine functions can though be seen as rather counterintuitive as every variable influences the function value dramatically. For all three studied problems, we exhibit the same trichotomy for the parameter |$|H|$| (⁠|${\textsf{IN}}$| is |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard, |${\textsf{IN}}_2$| is |${\textbf{para-}}{\textbf{DP}} $|-hard and the remaining are |${\textbf{FPT}} $|⁠). The parameter |$|V|$| always allows for FPT algorithms independent of the co-clone. Regarding |$|E|$|⁠, only the two size restricted variants are meaningful. For ‘|$\leq $|’ we achieve a pentachotomy between |${\textbf{FPT}} $|⁠, |$\textbf{W}[2]$|-complete, |$\textbf{W}[\textbf{P}] $|-complete, |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|- and |${\textbf{para-}}{\textbf{DP}} $|-hard. Whereas, for ‘=’, we achieve a hexachotomy additionally having |$\textbf{W}[1]$|-completeness for the essentially negative cases. These |$\textbf{W}[1]$|-hard cases are also surprising in the sense that for ‘|$\leq $|’ they are easy and |${\textbf{FPT}} $|⁠. Similarly, the same easy/hard-difference has been observed as well for |$|M|$| as the studied parameter. However, here, we distinguish between |${\textbf{para-}}{\textbf{NP}} $|-complete for ‘=’ and |${\textbf{FPT}} $| for ‘|$\leq $|’. The complete picture for ‘=’ and |$|M|$| is a tetrachotomy ranging through |${\textbf{FPT}} $|⁠, |${\textbf{para-}}{\textbf{NP}} $|-complete, |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard and |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-complete. Also, parameterized enumeration complexity [7, 8] is the next object of our investigations. Here, we already made some initial observations yielding |${\textbf{FPT-enum}}$|-algorithms for |$|V|$| and |${\textsf{BR}}$| as well as for |$|H|$| and |${\textsf{IE}}_2,{\textsf{IV}}_2,{\textsf{ID}}_2$| and |${\textsf{IL}}_2$|⁠. Such algorithms produce the whole set of solutions in |${\textbf{FPT}} $|-time. Furthermore, |${\textsf{IL}}_1$| even allows for |${\textbf{FPT}}$|-algorithms for any parameterization for the problem |$\textrm{ABD}(S, ks)$| (so it extends Corollary 3.10 in that way). Notice that in this paper, we did not require |$H\cap M$| to be empty as, for instance, Fellows et al. [21] assumed. All our proofs (e.g. Lemma 3.12) can be easily adapted in that direction. Furthermore, we believe that the |${\textbf{para-}}{\textbf{DP}} $|-hardness for |$|H|$| and |${\textsf{IN}}_2$| should be extendable to |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hardness but do not have a full proof yet. To close the outlook, one could attack the affine co-clones. Funding This work was supported by the German Research Foundation (DFG) under the project number ME 4279/1-2. References [1] M. Bauland , M. Mundhenk, T. Schneider, H. Schnoor, I. Schnoor, and H. Vollmer The tractability of model checking for LTL: the good, the bad, and the ugly fragments . ACM Transactions on Computational Logic (TOCL) , 12 , 13:1 – 13:28 , 2011 . Google Scholar Crossref Search ADS WorldCat [2] M. Bauland , T. Schneider, H. Schnoor, I. Schnoor, and H. Vollmer The complexity of generalized satisfiability for linear temporal logic . Logical Methods in Computer Science , 5 , 2009 . 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Parameterized complexity of abduction in Schaefer’s framework

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Abstract

Abstract Abductive reasoning is a non-monotonic formalism stemming from the work of Peirce. It describes the process of deriving the most plausible explanations of known facts. Considering the positive version, asking for sets of variables as explanations, we study, besides the problem of wether there exists a set of explanations, two explanation size limited variants of this reasoning problem (less than or equal to, and equal to a given size bound). In this paper, we present a thorough two-dimensional classification of these problems: the first dimension is regarding the parameterized complexity under a wealth of different parameterizations, and the second dimension spans through all possible Boolean fragments of these problems in Schaefer’s constraint satisfaction framework with co-clones (T. J. Schaefer. The complexity of satisfiability problems. In Proceedings of the 10th Annual ACM Symposium on Theory of Computing, May 1–3, 1978, San Diego, California, USA, R.J. Lipton, W.A. Burkhard, W.J. Savitch, E.P. Friedman, A.V. Aho eds, pp. 216–226. ACM, 1978). Thereby, we almost complete the parameterized complexity classification program initiated by Fellows et al. (The parameterized complexity of abduction. In Proceedings of the Twenty-Sixth AAAI Conference on Articial Intelligence, July 22–26, 2012, Toronto, Ontario, Canada, J. Homann, B. Selman eds. AAAI Press, 2012), partially building on the results by Nordh and Zanuttini (What makes propositional abduction tractable. Artificial Intelligence, 172, 1245–1284, 2008). In this process, we outline a fine-grained analysis of the inherent parameterized intractability of these problems and pinpoint their FPT parts. As the standard algebraic approach is not applicable to our problems, we develop an alternative method that makes the algebraic tools partially available again. 1 Introduction The framework of parameterized complexity theory yields a more fine-grained complexity analysis of problems than the classical worst-case complexity may achieve. Introduced by Downey and Fellows [17, 18], one associates problems with a specific parameterization, i.e. one studies the complexity of parameterized problems. Here, one aims to find parameters relevant for practice allowing to solve the problem by algorithms running in time |$f(k)\cdot n^{O(1)}$|⁠, where |$f$| is a computable function, |$k$| is the value of the parameter and |$n$| is the input length. Problems with such a running time are called fixed-parameter tractable (⁠|${\textbf{FPT}}$|⁠) and correspond to efficient computation in the parameterized setting. This is justified by the fact that parameters are usually slowly growing or even of constant value. Despite that, a different quality of runtimes is of the form |$n^{f(k)}$|⁠, which are obeyed by algorithms solving problems in the class |${\textbf{XP}}$|⁠. Comparing both classes with respect to the runtimes their problems allow to be solved in, of course, both runtimes are polynomial. However, for the first type, the degree of the polynomial is independent of the parameter’s value which is notable to observe. As a result, the second kind of runtimes is undesirable and usually tried to circumvented by locating different parameters. It is known that |${\textbf{FPT}}\subsetneq{\textbf{XP}}$| by diagonalization and also that a (presumably infinite) hierarchy of parameterized intractability in between these two classes exist: the so-called |$\textbf{W}$|-hierarchy, which is contained also in the class |$\textbf{W}[\textbf{P}] \subseteq{\textbf{XP}}$|⁠. These |$\textbf{W}$|-classes are regarded as a measure of intractability in the parameterized sense. Intuitively, showing |$\textbf{W}[1]$|-lower bounds corresponds to |${\textbf{NP}} $|-lower bounds in the classical setting. The limit of this hierarchy, the class |$\textbf{W}[\textbf{P}] $| is defined via nondeterministic machines that have at most |$h(k)\cdot \log n$| many nondeterministic steps, where |$h$| is a computable function, |$k$| the parameter’s value and |$n$| is the input length. Clearly, the process of human common-sense reasoning is non-monotonic, as adding further knowledge might decrease the number of deducible facts. As a result, non-monotonic logics became a well-established approach to investigate this kind of reasoning. One of the popular formalism in this area of research is abductive reasoning, which is an important concept in artificial intelligence as emphasized by Morgan [30] and Pole [36]. In particular, abduction is used in the process of medical diagnosis [24, 35] and thereby relevant for practice. Intuitively, abductive reasoning describes the process of deriving the most plausible explanations of known facts and originated from the work by Peirce [34]. Formally, one uses propositional formulas to model known facts in a knowledge base|${\textit{KB}}$| together with a set of manifestations|$M$| and a set of hypotheses|$H$|⁠. In this paper, |$H$| and |$M$| are sets of propositions as studied by Fellows et al. [21] as well as Eiter and Gottlob [20]. Formally, one tries to find a preferably small set of propositions |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|⁠. |$E$| is then called an explanation for |$M$|⁠. In this context, we distinguish three kinds of problems: the first just asks for such a very set |$E$| that fulfils these properties (⁠|$\textrm{ABD}$|⁠), the second tries to find a set of size less than or equal to a specific size (⁠|$\textrm{ABD}_\leq $|⁠) and the third one wants to spot a set of exactly a given size (⁠|$\textrm{ABD}_=$|⁠). Classically, |$\textrm{ABD}$| is complete for the second level of the polynomial hierarchy |${\boldsymbol{\varSigma _2^P}} $| [20] and its difficulty is very well understood [11, 16, 31, 42]. As a result, under reasonable complexity-theoretic assumptions, the problem is highly intractable posing the question in turn for sources of this complexity. In this direction, there exists research that aims to better understand the structure and difficulty of this problem, i.e. in the context of parameterized complexity. Here, Fellows et al. [21] initiated an investigation of possible parameters and classified CNF-induced fragments of the reasoning problems with respect to a multitude of parameters. The authors study the CNF-fragments Horn, Krom and DefHorn. They studied the parameterizations |$|M|$| (number of manifestations), |$|H|$| (number of hypotheses), |$|V|$| (number of variables) and |$|E|$| (number of explanations that is equivalent to their solution size |$k$|⁠) directly stemming from problem components, as well as the tree-width [39], and the size of the smallest vertex cover. In their classification, besides showing several |$ {\textbf{para-}}{\textbf{NP}} $|-/|$\textbf{W}[\textbf{P}]$|-complete/|$ {\textbf{FPT}} $| cases, they also focus on the existence of polynomial kernels and present a complete picture regarding their CNF-classes. Universal algebra yields a systematic way to rigorously classify fragments of a problem induced by restricting its Boolean connectives. This technique is built around Post’s lattice [37] which bases on the notion of (co-)clones. Intuitively, given a set of Boolean functions |$B$|⁠, the clone of |$B$| is the set of functions that are expressible by compositions of functions from |$B$| (plus introducing fictive variables). The most prominent result under this approach is the dichotomy theorem of Lewis [26], which classifies propositional satisfiability into polynomial-time solvable cases and intractable ones depending merely on the existence of specific Boolean operators. This approach has been followed many times in a wealth of different contexts [2, 3, 9, 14, 28, 29, 38] as well as in the context of abduction itself [12, 31]. Interestingly, in the scope of constraint satisfaction problems, the investigation of co-clones (or relational clones) allows one to proceed a similar kind of classification (see, e.g. the work by Nordh and Zanuttini [31]). The reason for that lies in the concept of invariance of relations under some function |$f$| (one defines this property via polymorphisms where |$f$| is applied component-wisely to the columns of the relation). In view of this, Post’s lattice supplies a similar lattice, now for sets of relations that are invariant under respective functions. With respect to constraint satisfaction, the most prominent classification is due to Schaefer [40] who similarly divides the constraint satisfaction problem restricted to co-clones into polynomial-time solvable and |$ {\textbf{NP}} $|-complete cases. The algebraic approach has been successfully applied to abduction by Nord and Zanuttini [31]. For the problems that we consider, it is less obvious how to use the algebraic tools: the standard trick to obtain reductions preserves the existence of explanations, but not their size. Due to this, we develop an alternative method that makes the algebraic tools partially available again (see Section 3.1). Much in the vein of Schaefer’s classification, we present a thorough study directly pinpointing those restrictions of the abductive reasoning problem, which yield efficiency under the parameterized approach. In a sense, we present an almost complete picture that has been initiated by Fellows et al. [21] except for some minor cases around the affine co-clones. Their classification is covered by our study now, as Horn cases correspond to the co-clones below |$ {\textsf{IE}}_2$|⁠, DefHorn conforms |$ {\textsf{IE}}_1$| and Krom matches with |${\textsf{ID}}_2$|⁠. The motivation of our research is to draw a finer line than Fellow et al. did and to present complete picture with respect to all possible constraint languages now. From this classification, we draw some surprising results. Regarding the essentially negative cases for the parameter |$|M|$|⁠, |$\textrm{ABD}_=$| is |${\textbf{para-}}{\textbf{NP}} $|-complete, whereas |$\textrm{ABD}_\leq $| is |${\textbf{FPT}} $|⁠. Also for this parameter, |${\textsf{IE}}_1$| and |${\textsf{IE}}$| are hard for |$\textrm{ABD}_=$| and |$\textrm{ABD}_\leq $| (both |${\textbf{para-}}{\textbf{NP}} $|-complete), but |$\textrm{ABD}$| is |${\textbf{FPT}} $|⁠. Regarding |$|E|$| as parameterization, the behaviour is similarly unexpected for the essentially negative cases: |${\textbf{FPT}} $| for |$\textrm{ABD}_\leq $| versus |$\textbf{W}[1]$|-hardness for |$\textrm{ABD}_=$|⁠. Another interesting remark regarding the essentially negative languages is that the problem |$\textrm{ABD}_={(|M|)}$| is harder than |$\textrm{ABD}_\leq{(|E|)}$| (⁠|${\textbf{para-}}{\textbf{NP}} $| vs |$\textbf{W} [2]$|⁠), whereas |$\textrm{ABD}_\leq{(|M|)}$| is easier than |$\textrm{ABD}_\leq{(|E|)}$| (⁠|$\textbf{W}[1]$| vs |$\textbf{W}[2]$|⁠). For the parameters |$|V|$| as well as |$|H|$| the classifications for all three problems are the same. Figure 1 shows our results for all problems and parameterizations in a single picture. Figure 1. Open in new tabDownload slide Complexity landscape of abductive reasoning with respect to the studied parameters |$|M|,|H|,|V|,|E|$|⁠. White colouring means unclassified. |$\textrm{ABD}_\star $| means same result for all three variants. Figure 1. Open in new tabDownload slide Complexity landscape of abductive reasoning with respect to the studied parameters |$|M|,|H|,|V|,|E|$|⁠. White colouring means unclassified. |$\textrm{ABD}_\star $| means same result for all three variants. The paper is organized as follows. We first recall the basic concepts from parameterized complexity theory, the notion of co-clones and formalize the abduction problem. Section 3 includes our complexity results, we first prove an important result in Section 3.1. Then we give some general intractable as well as tractable cases that are true under every parameterization. This is followed by subsections discussing individual parameters with Theorems summarizing our results. We conclude in Section 4 with some interesting remarks and open cases. 2 Preliminaries We require standard notions from classical complexity theory [33]. We encounter the classical complexity classes P, NP, |${\textbf{DP}} = \{A \setminus B \mid A,B\in{\textbf{NP}} \}$|⁠, |${\textbf{co}}{\textbf{NP}} $|⁠, |${\boldsymbol{\varSigma _2^P}} ={\textbf{NP}} ^{\textbf{NP}} $| and their respective completeness notions, employing polynomial time many-one reductions (⁠|$\leq ^{\textbf{P}}_m$|⁠). Parameterized Complexity Theory. A parameterized problem (PP) |$P\subseteq \varSigma ^*\times \mathbb N$| is a subset of the crossproduct of an alphabet and the natural numbers. For an instance |$(x,k)\in \varSigma ^*\times \mathbb N$|⁠, |$k$| is called the (value of the) parameter. A parameterization is a polynomial-time computable function that maps a value from |$x\in \varSigma ^*$| to its corresponding |$k\in \mathbb N$|⁠. The problem |$P$| is said to be fixed-parameter tractable (or in the class |${\textbf{FPT}} $|⁠) if there exists a deterministic algorithm |$\mathcal A$| and a computable function |$f$| such that for all |$(x,k)\in \varSigma ^*\times \mathbb N$|⁠, algorithm |$\mathcal A$| correctly decides the membership of |$(x,k)\in P$| and runs in time |$f(k)\cdot |x|^{O(1)}$|⁠. The problem |$P$| belongs to the class |${\textbf{XP}} $| if |$\mathcal A$| runs in time |$|x|^{f(k)}$|⁠. There exists a hierarchy of complexity classes in between |${\textbf{FPT}} $| and |${\textbf{XP}} $|⁠, which is called |$\textbf{W}$|-hierarchy (for details see the textbook of Flum and Grohe [22]). We will make use of the classes |$\textbf{W}[1]$| and |$\textbf{W}[2]$|⁠. Complete problems characterizing these classes are introduced later in Proposition 2.4. Also, we work with classes that can be defined via a precomputation on the parameter. Definition 2.1 Let |$\mathcal C$| be any complexity class. Then |${\textbf{para-}} \mathcal C$| is the class of all PPs |$P\subseteq \varSigma ^*\times \mathbb N$| such that there exists a computable function |$\pi \colon \mathbb N\to \varDelta ^*$| and a language |$L\in \mathcal C$| with |$L\subseteq \varSigma ^*\times \varDelta ^*$| such that for all |$(x,k)\in \varSigma ^*\times \mathbb N$| we have that |$(x,k)\in P \Leftrightarrow (x,\pi (k))\in L$|⁠. Notice that |${\textbf{para-}} \textbf{P}={\textbf{FPT}} $|⁠. The complexity classes |$\mathcal{C}\in{\{\, {\textbf{NP}} ,{\textbf{co}}{\textbf{NP}} , {\textbf{DP}} ,{\boldsymbol{\varSigma _2^P}} \,\}} $| are used in the |${\textbf{para-}} \mathcal C$| context by us. Let |$c\in \mathbb N$| and |$P\subseteq \varSigma ^*\times \mathbb N$| be a PP, then the |$c$|-slice of|$P$|⁠, written as |$P_c$| is defined as |$P_c:=\{\,(x,k)\in \varSigma ^*\times \mathbb N\mid k=c\,\}$|⁠. Notice that |$P_c$| is a classical problem then. Observe that, regarding our studied complexity classes, showing membership of a PP |$P$| in the complexity class |${\textbf{para-}} \mathcal C$|⁠, it suffices to show that each slice |$P_c\in \mathcal C$|⁠. Definition 2.2 Let |$P\subseteq \varSigma ^*\times \mathbb N$| and |$Q\subseteq \varGamma ^*\times \mathbb N$| be two PPs. One says that |$P$| is fpt-reducible to |$Q$|⁠, if there exists an fpt-computable function |$f\colon \varSigma ^*\times \mathbb N\to \varGamma ^*\times \mathbb N$| such that – for all |$(x,k)\in \varSigma ^*\times \mathbb N$| we have that |$(x,k)\in P\Leftrightarrow f(x,k)\in Q$|⁠, there exists a computable function |$g\colon \mathbb N\to \mathbb N$| such that for all |$(x,k)\in \varSigma ^*\times \mathbb N$| and |$f(x,k)=(x^{\prime},k^{\prime})$| we have that |$k^{\prime}\leq g(k)$|⁠. Propositional Logic. We assume familiarity with propositional logic. A literal is a variable |$x$| or its negation |$\neg x$|⁠. A clause is a disjunction of literals and a term is a conjunction of literals. We denote by |$\textrm{var}(\varphi )$| the variables of a formula |$\varphi $|⁠. Analogously, for a set of formulas |$F$|⁠, |$\textrm{var}(F)$| denotes |$\bigcup _{\varphi \in F}\textrm{var}(\varphi )$|⁠. We identify finite |$F$| with the conjunction of all formulas from |$F$|⁠, i.e. |$\bigwedge _{\varphi \in F} \varphi $|⁠. A mapping |$\sigma \colon \textrm{var}(\varphi ) \mapsto \{0,1\}$| is called an assignment to the variables of |$\varphi $|⁠. A model of a formula |$\varphi $| is an assignment to |$\textrm{var}(\varphi )$| that satisfies |$\varphi $|⁠. The weight of an assignment |$\sigma $| is the number of variables |$x$| such that |$\sigma (x)=1$|⁠. For two formulas |$\psi , \varphi $| we write |$\psi \models \varphi $| if every model of |$\psi $| also satisfies |$\varphi $|⁠. A formula is positive (resp., negative) if every literal appears positively (negatively) and a negation symbol appears only in front of a variable. The class of all propositional formulas is denoted by |${\textit{PROP}}$|⁠. Occasionally, in this paper, we will consider special subclasses of formulas, namely $$\begin{align*} \varGamma_{0, d} & = {\{\, \ell_1\land\ldots\land \ell_c \mid \ell_1,\ldots, \ell_c\ \textrm{are literals and}\ c\leq d \,\}},\\ \varDelta_{0, d} & = {\{\, \ell_1\lor\ldots\lor \ell_c \mid \ell_1,\ldots, \ell_c\ \textrm{are literals and}\ c\leq d \,\}},\\ \varGamma_{t, d} & = \left\{\,\bigwedge\limits_{i\in I} \alpha_i \,\middle|\, \alpha_i \in \varDelta_{t-1,d}\ \textrm{for}\ i \in I\, \right\}, \varDelta_{t, d} = \left\{\,\bigvee\limits_{i\in I} \alpha_i \,\middle|\, \alpha_i \in \varGamma_{t-1,d},\ i\ \in\ I\, \right\}. \end{align*}$$ Finally, |$\varGamma ^{+}_{t,d}$| (resp. |$\varGamma ^{-}_{t,d}$|⁠) denote the class of all positive (negative) formulas in |$\varGamma _{t,d}$|⁠. Example 2.3 Let |$\phi = \bigwedge _{i\leq m}(\neg x_{i,1} \lor \cdots \lor \neg x_{i,n_i})$| for |$1\leq n_i\leq d$| and |$d,m\in \mathbb N$|⁠. That is, |$\phi $| is a conjunction of the clauses containing negative literals. Then |$\phi \in \varGamma _{1,d}$|⁠, the so-called |$d$|-CNF. Note also that |$\phi $| is an |${{\textsf{IS}^{d}_{1}}}$|-formula using only negative clauses. We will often reduce a problem instance to (and from) a parameterized weighted satisfiability problem for propositional formulas. This problem is defined below. Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Open in new tab Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Problem: . |${\textrm{p-WSAT}(\varGamma _{t,d})}$| . Input: A |$\varGamma _{t,d}$|-formula |$\alpha $| over variables |$V$| with |$t, d \geq 1 $| and |$k\in \mathbb{N}$|⁠. Parameter: |$k$|⁠. Question: Is there a satisfying assignment for |$\alpha $| of weight |$k$|? Open in new tab Two similarly defined problems are |${\textrm{p-WSAT}(\varGamma ^+_{t,1})}$| and |${\textrm{p-WSAT}(\varGamma ^-_{t,1})}$| where an instance |$\alpha $| comes from classes |$ \varGamma ^{+}_{t,1}$| (resp. |$\varGamma ^{-}_{t,1}$|⁠). The classes of the |$\textbf{W}$|-hierarchy can be defined in terms of these problems as proved by Downey and Fellows [22]. Proposition 2.4 ([22]). The following problems are |$\textbf{W}[t]$|-complete for each |$t \geq 1$|⁠, under |$\leq ^{\textbf{FPT}}$|-reductions: – |${\textrm{p-WSAT}(\varGamma ^+_{t,1})} $| if t is even, |${\textrm{p-WSAT}(\varGamma ^-_{t,1})} $| if t is odd, |${\textrm{p-WSAT}(\varGamma _{t,d})}$| for every |$t$| and |$d\geq 1$|⁠. Constraints and |${S}$|-formulas. A logical relation of arity |$k$| is a relation |$R \subseteq \{0,1\}^k$|⁠. A constraint is a formula |$R(x_1, \dots , x_k)$|⁠, where |$R$| is a logical relation of arity |$k$| and the |$x_i$|’s are (not necessarily distinct) variables. An assignment |$\sigma $| to the |$x_i$|’s satisfies the constraint if |$(\sigma (x_1), \dots , \sigma (x_k)) \in R$|⁠. A constraint language|${S}$| is a finite set of logical relations. An |${S}$|-formula|$\varphi $| is a conjunction of constraints built upon logical relations only from |${S}$| and accordingly can be seen as a quantifier-free first-order formula. An assignment |$\sigma $| is called a model of |$\varphi $| if |$\sigma $| satisfies all constraints in |$\varphi $| simultaneously. Whenever an |${S}$|-formula or constraint is logically equivalent to a single clause or term, we treat it as such. Definition 2.5 1. The set |$\left \langle{S} \right \rangle $| is the smallest set of relations that contains |${S}$|⁠, the equality constraint, =, and which is closed under primitive positive first order definitions, i.e. if |$\phi $| is an |${S} \cup \{=\}$|-formula and |$R(x_1, \dots , x_n) \equiv \exists y_1 \dots \exists y_l \phi (x_1, \dots , x_n,y_1, \dots , y_l)$|⁠, then |$R \in \left \langle{S} \right \rangle $|⁠. In other words, |$\left \langle{S} \right \rangle $| is the set of relations that can be expressed as an |${S} \cup \{=\}$|-formula with existentially quantified variables. 2. The set |$\left \langle{S} \right \rangle _{\neq }$| is the set of relations that can be expressed as an |${S}$|-formula with existentially quantified variables (no equality relation is allowed). The set |$\left \langle{S} \right \rangle $| is called a relational clone or co-clone with base|${S}$| [4]. Notice that for a co-clone |${\textsf{C}}$| and a constraint language |${S}$| the statements |${S} \subseteq{\textsf{C}}$|⁠, |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}$| and |$\left \langle{S} \right \rangle _{\neq } \subseteq{\textsf{C}}$| are equivalent. Throughout the text, we refer to different types of Boolean relations and corresponding co-clones following Schaefer’s terminology [40]. For an overview of co-clones and bases, see Table 1. Note that |$\left \langle{S} \right \rangle _{\neq } \subseteq \left \langle{S} \right \rangle $| by definition. The other direction does not hold in general. However, if |$(x = y) \in \left \langle{S} \right \rangle _{\neq }$|⁠, then |$\left \langle{S} \right \rangle _{\neq } = \left \langle{S} \right \rangle $|⁠. Table 1. Overview of bases [4] and clause descriptions [31] for co-clones, where EVEN|$^4$| = |$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus 1$|⁠. Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Open in new tab Table 1. Overview of bases [4] and clause descriptions [31] for co-clones, where EVEN|$^4$| = |$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus 1$|⁠. Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Co-clone . Base . Clause type . Name/indication . |${\textsf{BR}}$| (⁠|${\textsf{II}}_2$|⁠) 1-IN-3 = |$\{001, 010, 100\}$| All clauses All Boolean relations |${\textsf{II}}_1$| |$x \lor (y \oplus z)$| At least one positive literal per clause 1-valid |${\textsf{II}}_0$| DUP, |$x \rightarrow y$| At least one negative literal per clause 0-valid |${\textsf{II}}$| EVEN|$^4$|⁠, |$x \rightarrow y$| At least one negative and one positive literal per clause 1- and 0-valid |${\textsf{IN}}_2$| NAE = |$\{0,1\}^3 \setminus \{000,111\}$| Cf. previous column Complementive |${\textsf{IN}}$| DUP = |$\{0,1\}^3 \setminus \{101, 010\}$| Cf. previous column Complementive and 1- and 0-valid |${\textsf{IE}}_2$| |$x \land y \rightarrow z, x, \neg x$| Clauses with at most one positive literal Horn |${\textsf{IE}}_1$| |$x \land y \rightarrow z, x$| Clauses with exactly one positive literal Definite Horn |${\textsf{IE}}_0$| |$x \land y \rightarrow z, \neg x$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2, (\neg x_1 \lor \dots \lor \neg x_n), n \geq 1$| Horn and 0-valid |${\textsf{IE}}$| |$x \land y \rightarrow z$| |$(x_1 \lor \neg x_2 \lor \dots \lor \neg x_n), n\geq 2$| Horn and 1- and 0-valid |${\textsf{IV}}_2$| |$x \lor y \lor \neg z, x, \neg x$| Clauses with at most one negative literal DualHorn |${\textsf{IV}}_1$| |$x \lor y \lor \neg z, x$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2, (x_1 \lor \dots \lor x_n), n \geq 1$| DualHorn and 1-valid |${\textsf{IV}}_0$| |$x \lor y \lor \neg z, \neg x$| Clauses with exactly one negative literal Definite dualHorn |${\textsf{IV}}$| |$x \lor y \lor \neg z$| |$(\neg x_1 \lor x_2 \lor \dots \lor x_n), n\geq 2$| DualHorn and 1- and 0-valid |${\textsf{IL}}_2$| EVEN|$^4$|⁠, |$x$|⁠, |$\neg x$| All affine clauses (all linear equations) Affine |${\textsf{IL}}_1$| EVEN|$^4$|⁠, |$x$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n\geq 0, a = n$| (mod 2) Affine and 1-valid |${\textsf{IL}}_0$| EVEN|$^4$|⁠, |$\neg x$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n\geq 0$| Affine and 0-valid |${\textsf{IL}}_3$| EVEN|$^4$|⁠, |$x \oplus y$| |$(x_1 \oplus \dots \oplus x_n = a)$|⁠, |$n$| even, |$a \in \{0,1\}$| - |${\textsf{IL}}$| EVEN|$^4$| |$(x_1 \oplus \dots \oplus x_n = 0)$|⁠, |$n$| even Affine and 1- and 0-valid |${\textsf{ID}}_2$| |$x \oplus y, x \rightarrow y$| Clauses of size 1 or 2 Bijunctive, KROM, 2CNF |${\textsf{ID}}_1$| |$x \oplus y, x, \neg x$| Affine clauses of size 1 or 2 2-affine |${\textsf{ID}}$| |$x \oplus y$| Affine clauses of size 2 Strict 2-affine |${\textsf{IM}}_2$| |$x \rightarrow y, x, \neg x$| |$(x_1 \rightarrow x_2), (x_1), (\neg x_1)$| Implicative |${\textsf{IM}}_1$| |$x \rightarrow y, x$| |$(x_1 \rightarrow x_2), (x_1)$| Implicative and 1-valid |${\textsf{IM}}_0$| |$x \rightarrow y, \neg x$| |$(x_1 \rightarrow x_2), (\neg x_1)$| Implicative and 0-valid |${\textsf{IM}}$| |$x \rightarrow y$| |$(x_1 \rightarrow x_2)$| Implicative and 1- and 0-valid |${\textsf{IS}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| IHS-B- |${\textsf{IS}^{k}_{10}}$| Cf. next column |$(x_1), (x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| IHS-B- of width |$k$| |${\textsf{IS}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Essentially negative |${\textsf{IS}^{k}_{12}}$| Cf. next column |$(x_1), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially negative of width |$k$| |${\textsf{IS}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), n \geq 0$| - |${\textsf{IS}^{k}_{11}}$| Cf. next column |$(x_1 \rightarrow x_2), (\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0$| - |${\textsf{IS}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), n \geq 0, (x_1 = x_2)$| Negative |${\textsf{IS}^{k}_{1}}$| Cf. next column |$(\neg x_1 \lor \dots \lor \neg x_n), k \geq n \geq 0, (x_1 = x_2)$| Negative of width |$k$| |${\textsf{IS}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| IHS-B+ |${\textsf{IS}^{k}_{00}}$| Cf. next column |$(\neg x_1), (x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| IHS-B+ of width |$k$| |${\textsf{IS}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Essentially positive |${\textsf{IS}^{k}_{02}}$| Cf. next column |$(\neg x_1), (x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Essentially positive of width |$k$| |${\textsf{IS}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), n \geq 0$| - |${\textsf{IS}^{k}_{01}}$| Cf. next column |$(x_1 \rightarrow x_2), (x_1 \lor \dots \lor x_n), k \geq n \geq 0$| - |${\textsf{IS}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), n \geq 0, (x_1 = x_2)$| Positive |${\textsf{IS}^{k}_{0}}$| Cf. next column |$(x_1 \lor \dots \lor x_n), k \geq n \geq 0, (x_1 = x_2)$| Positive of width |$k$| |${\textsf{IR}}_2$| |$x_1, \neg x_2$| |$(x_1), (\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}_1$| |$x_1$| |$(x_1), (x_1 = x_2)$| - |${\textsf{IR}}_0$| |$\neg x_1$| |$(\neg x_1), (x_1 = x_2)$| - |${\textsf{IR}}$| (⁠|${\textsf{IBF}}$|⁠) |$\emptyset $| |$(x_1 = x_2)$| - Open in new tab Abduction. An instance of the abduction problem for |${S}$|-formulas is given by |$\langle V, H, M, {\textit{KB}} \rangle $|⁠, where |$V$| is the set of variables, |$H$| is the set of hypotheses, |$M$| is the set of manifestations and |${\textit{KB}}$| is the knowledge base (or theory) built upon variables from |$V$|⁠. A knowledge base |${\textit{KB}}$| is a set of |${S}$|-formulas that we assimilate with the conjunction of all formulas it contains. We define the following abduction problems for |$S$|-formulas. Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}(S,k)$|—the abductive reasoning problem for |${S}$|-formulas parameterized by |$k$| . Input: |$\langle V, H, M, {\textit{KB}}, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Similarly, the problem |$\textrm{ABD}(S)$| is the classical pendant of |$\textrm{ABD}(S,k)$|⁠. Additionally, we consider size restrictions for a solution and define the following problems. Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Problem: . |$\textrm{ABD}_\leq(S,k)$| . Input: |$\langle V, H, M, {\textit{KB}}, s, k \rangle $|⁠, where |${\textit{KB}}$| is a set of |${S}$|-formulas, |$H, M$| are each set of propositions and |$V=\textrm{var}(H)\cup \textrm{var}(M)\cup \textrm{var}({\textit{KB}})$|⁠, and |$s\in \mathbb N$|⁠. Parameter: |$k$|⁠. Question: Is there a set |$E\subseteq H$| with |$|E|\leq s$| such that |$E\land{\textit{KB}}$| is satisfiable and |$E\land{\textit{KB}}\models M$|? Open in new tab Analogously, |$\textrm{ABD}_=(S,k)$| requires the size of |$E$| to be exactly |$s$| and |$\textrm{ABD}_=(S), \textrm{ABD}_\leq(S)$| are the classical counterparts. Notice that, for instance, in cases where the parameter is the size of solutions, then |$s=k$|⁠. Example 2.6 Sitting in a train you realize that it is still not moving even though the clock suggests it should be. You start reasoning about it. Either some door is open, the train has delayed or that engine has failed. This form of reasoning is called abductive reasoning. Having some additional information that the operator of train usually announces in case the train is delayed or engine has failed, you deduce that some door must be opened and that train will start moving soon when all the doors are closed. Formally, one is interested in an explanation for the observed event (manifestation) |${\{\, \texttt{stop} \,\}} $|⁠. The knowledge base includes the following statements: |$-$||$\neg \texttt{moving} \leftrightarrow \texttt{stop} \qquad\qquad\qquad\qquad \ \ \ \ \ \ -\texttt{trainDelayed} \rightarrow \texttt{newTime},$| |$\neg \texttt{announcement}$|⁠, |$\texttt{moving} \rightarrow \texttt{time}, \qquad\qquad\qquad\qquad\qquad-\ (\texttt{engineFailed} \lor \texttt{trainDelayed}\ \lor$| |$\texttt{engineFailed}\rightarrow \texttt{announcement},\qquad\quad \texttt{doorOpen} )\rightarrow \texttt{stop}$|⁠, Then the set of hypotheses |${\{\, \texttt{time}, \texttt{doorOpen}, \texttt{announcement} \,\}} $| has an explanation, namely, |${\{\, \texttt{doorOpen} \,\}} $|⁠. On the other hand, |${\{\, \texttt{time} \,\}} $| does not explain the event |${\{\, \texttt{stop} \,\}} $|⁠, whereas |${\{\, \texttt{announcement} \,\}} $| is not consistent with the knowledge base. Consequently, an explanation of size |$1$| exists. There also exists an explanation of size |$2$| since |${\{\, \texttt{time}, \texttt{doorOpen} \,\}} $| is consistent with |${\textit{KB}}$| and explains |$M$|⁠. Note that having the set of hypotheses |${\{\, \texttt{engineFailed}, \texttt{doorOpen} \,\}} $| facilitates only one explanation of size |$1$|⁠, namely, |${\{\, \texttt{doorOpen} \,\}} $|⁠, even though the hypotheses set has size |$2$|⁠. Let |${\textrm{SAT}}$| and |${\textrm{IMP}}$| denote the classical satisfiability and implication problems. Given a constraint language |${S}$| then an instance of |${\textrm{SAT}}({S})$| is an |${S}$|-formula |$\varphi $| and the question is whether there exists a satisfying assignment for |$\varphi $|⁠. On the other hand, an instance of |${\textrm{IMP}}({S})$| is |$(\phi ,\psi )$| such that |$\phi , \psi $| are two |${S}$|-formulas and the question is whether |$\phi \models \psi $|⁠. We have the following observation regarding the classical |${\textrm{SAT}}$| and |${\textrm{IMP}}$| problems. Proposition 2.7 ([40, 41]). Let |${S}$| be a constraint language such that |${S} \subseteq{\textsf{C}}$| where |${\textsf{C}} \in{\{\, {\textsf{ID}}_2,{\textsf{IV}}_2, {\textsf{IE}}_2, {\textsf{IL}}_2 \,\}} $|⁠. Then |${\textrm{SAT}}({S})$| and |${\textrm{IMP}}({S})$| are both in |${\textbf{P}} $|⁠. 3 Complexity results for abductive reasoning We begin by presenting a number of technical expressivity results that allow us in the sequel to prove a crucial property for the whole classification endeavour (Lemma 3.3). 3.1 Base independence The idea of lemma 3.2 is to express equality by some other construction, we need the following proposition for the proof. ([6]) . Proposition 3.1 Let |${S}$| be a constraint language. The following is true: 1. If |${S}$| is not 1-valid, not 0-valid and complementive, then |$(x\neq y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.6.1]. 2. If |${S}$| is not 1-valid, not 0-valid, and not complementive, then |$(x\land \neg y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.6.3]. 3. If |${S}$| is 1-valid, 0-valid and not trivial, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.7]. 4. If |${S}$| is 1-valid, not 0-valid and not essentially positive, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.8.1]. 5. If |${S}$| is not 1-valid, 0-valid and not essentially negative, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| [6, Lem. 4.8.2]. Lemma 3.2 Let |${S}$| be a constraint language. If |${S}$| is not essentially negative and not essentially positive, then |$(x=y) \in \left \langle{S} \right \rangle _{\neq }$| and |$\left \langle{S} \right \rangle = \left \langle{S} \right \rangle _{\neq }$|⁠. Proof. For constraint languages (relations) that are Horn (⁠|${\textsf{IE}}_2$|⁠), dualHorn (⁠|${\textsf{IV}}_2$|⁠), essentially negative (⁠|${\textsf{IS}_{12}}$|⁠) or essentially positive (⁠|${\textsf{IS}_{02}}$|⁠) we use the following characterizations by polymorphisms (see, e.g. [15]). The binary operations of conjunction, disjunction and negation are applied coordinate-wise. |$R$| is Horn if and only if |$m_1,m_2 \in R$| implies |$m_1 \land m_2 \in R$|⁠. |$R$| is dualHorn if and only if |$m_1,m_2 \in R$| implies |$m_1 \lor m_2 \in R$|⁠. |$R$| is essentially negative if and only if |$m_1,m_2,m_3 \in R$| implies |$m_1 \land (m_2 \lor \neg m_3) \in R$|⁠. |$R$| is essentially positive if and only if |$m_1,m_2,m_3 \in R$| implies |$m_1 \lor (m_2 \land \neg m_3) \in R$|⁠. In order to complete the proof of the lemma, we make a case distinction according to whether |${S}$| is 1- and/or 0-valid. 1-valid and 0-valid. This case follows immediately from Prop. 3.1, third item. 1-valid and not 0-valid. Follows immediately from Prop. 3.1, fourth item. not 1-valid and 0-valid. Follows immediately from Prop. 3.1, fifth item. not 0-valid and not 1-valid. We make another case distinction according to whether |${S}$| is Horn and/or dualHorn. not Horn and not dualHorn. It suffices to show that inequality |$(x \neq y)$| can be expressed, since |$(x=y) \equiv \exists z(x\neq z) \land (z\neq y)$|⁠. If |${S}$| is complementive, we obtain by Prop. 3.1, first item, that |$(x\neq y) \in \left \langle{S} \right \rangle _{\neq }$|⁠. Therefore suppose now that |${S}$| is not complementive.Let |$R$| be a relation that is not Horn. Then there are |$m_1,m_2 \in R$| such that |$m_1 \land m_2 \notin R$|⁠. For |$i,j \in \{0,1\}$|⁠, set |$V_{i,j} = \{x \mid x \in V,\ m_1(x) = i,\ m_2(x) = j\}$|⁠. Observe that the sets |$V_{0,1}$| and |$V_{1,0}$| are nonempty (otherwise |$m_1 = m_1 \land m_2$| or |$m_2 = m_1 \land m_2$|⁠, a contradiction). Denote by |$C$| the |$\{R\}$|-constraint |$C = R(x_1, \dots , x_k)$|⁠. Set $$\begin{align*} &M_1(u,x,y,v) = C[V_{0,0}/u, V_{0,1}/x, V_{1,0}/y, V_{1,1}/v].\end{align*}$$ It contains |$\{0011, 0101\}$| (since |$m_1, m_2 \in R$|⁠), but it does not contain |$0001$| (since |$m_1 \land m_2 \notin R$|⁠). Let |$R$| be a relation that is not dualHorn. Then there are |$m_3,m_4 \in R$| such that |$m_3 \lor m_4 \notin R$|⁠. For |$i,j \in \{0,1\}$|⁠, set |$V^{\prime}_{i,j} = \{x \mid x \in V,\ m_3(x) = i,\ m_4(x) = j\}$|⁠. Observe that the sets |$V^{\prime}_{0,1}$| and |$V^{\prime}_{1,0}$| are nonempty (otherwise |$m_3 = m_3 \lor m_4$| or |$m_4 = m_3 \lor m_4$|⁠, a contradiction). Set |$M_2(u,x,y,v) = C[V^{\prime}_{0,0}/u, V^{\prime}_{0,1}/x, V^{\prime}_{1,0}/y, V^{\prime}_{1,1}/v]$|⁠. It contains |$\{0011, 0101\}$| (since |$m_3, m_4 \in R$|⁠), but it does not contain |$0111$| (since |$m_3 \lor m_4 \notin R$|⁠). Finally consider the |$\{R, (t \land \neg f)\}$|-formula $$\begin{align*} &M(f,x,y,t) = M_1(f,x,y,t) \land M_2(f,x,y,t) \land (t \land \neg f).\end{align*}$$ One verifies that it is equivalent to |$(x\neq y) \land (t \land \neg f)$|⁠. Due to Prop. 3.1, second item, |$(t \land \neg f)$| is expressible as an |$S$|-formula, and therefore so is |$M(f,x,y,t)$|⁠. Since |$\exists t,f\,\! M(f,x,y,t)$| is equivalent to |$(x\neq y)$|⁠, we obtain |$(x\neq y) \in \left \langle{S} \right \rangle _{\neq }$|⁠. Horn. Let |$R$| be a relation that is not essentially negative, but Horn. Then there are |$m_1,m_2,m_3 \in R$| such that |$m_4:= m_1 \land (m_2 \lor \neg m_3) \notin R$|⁠. Since |$R$| is Horn, |$m_5:= m_1 \land m_2 \in R$|⁠. For |$i,j,k \in \{0,1\}$|⁠, set |$V_{i,j,k} = \{x \mid x \in V,\ m_1(x) = i,\ m_2(x) = j,\ m_3(x) = k\}$|⁠. Observe that the sets |$V_{1,0,0}$| and |$V_{1,0,1}$| are nonempty (otherwise |$m_5 = m_4$| or |$m_1 = m_4$|⁠, a contradiction). Denote by |$C$| the |$\{R\}$|-constraint |$C = R(x_1, \dots , x_k)$|⁠. Set |$M(f,x,y,t) = C[V_{0,0,0}/f, V_{0,0,1}/f, V_{0,1,0}/f, V_{0,1,1}/f, V_{1,0,0}/x, V_{1,0,1}/y, V_{1,1,0}/t, V_{1,1,1}/t]$|⁠. It contains |$\{00001111,00000011\}$| (since |$m_1, m_5 \in R$|⁠), but it does not contain |$00001011$| (since |$m_4 \notin R$|⁠). Finally consider the |$\{R, (t\land \neg f)$|-formula $$\begin{align*} &M^{\prime}(f,x,y,t) = M(f,x,y,t) \land M(f,y,x,t) \land (t \land \neg f)\end{align*}$$ One verifies that it contains |$\{0111, 0001\}$| but not |$0101$| and neither |$0011$|⁠. Therefore it is equivalent to |$(x=y) \land (t \land \neg f)$|⁠. Due to Prop. 3.1, second item, |$(t \land \neg f)$| is expressible as an |$S$|-formula, and therefore so is |$M^{\prime}(f,x,y,t)$|⁠. Since |$\exists t,f\,\! M^{\prime}(f,x,y,t)$| is equivalent to |$(x = y)$|⁠, we obtain |$(x = y) \in \left \langle{S} \right \rangle _{\neq }$|⁠. dualHorn. Analogously to the Horn case, using the property that |${S}$| is not essentially positive, but dualHorn. The following property is crucial for presented results in the course of this paper. It supplies generalized upper as well as lower bounds (independence of the base of a co-clone), as long as the constraint language is not essentially negative and not essentially positive. The proof idea is to implement the previous lemma. Lemma 3.3 Let |${S}, {S}^{\prime}$| be two constraint languages such that |${S}^{\prime}$| is neither essentially positive nor essentially negative. Let |$\textrm{ABD}_* \in \{\textrm{ABD}, \textrm{ABD}_=, \textrm{ABD}_\leq\}$|⁠. If |${S} \subseteq \left \langle{S}^{\prime} \right \rangle $|⁠, then |$\textrm{ABD}_*(S) \leq^{\textbf{P}}_m \textrm{ABD}_*(S^\prime)$|⁠. Proof. We may consider |${\textit{KB}}$| as a single |${S}$|-formula. We will transform |${\textit{KB}}$| into a corresponding |${S}^{\prime}$|-formula by replacing every |${S}$|-constraint by a corresponding |${S}^{\prime}$|-formula. For this purpose we first construct a look-up table mapping any |$R \in{S}$| to an |${S}^{\prime}$|-formula |$F_R$| as follows. Since |$R \in{S} \subseteq \left \langle{S}^{\prime} \right \rangle $|⁠, and by Lemma 3.2|$\left \langle{S}^{\prime} \right \rangle = \left \langle{S}^{\prime} \right \rangle _{\neq }$|⁠, we have that |$R \in \left \langle{S}^{\prime} \right \rangle _{\neq }$|⁠. Thus, we have by definition an |${S}^{\prime}$|-formula |$\phi $| such that |$R(x_1, \dots , x_n) \equiv \exists y_1 \dots \exists y_m \phi (x_1, \dots , x_n,y_1, \dots , y_m)$|⁠, where we can assume the |$x_i$|’s and |$y_i$|’s to be |$n+m$| distinct variables. We obtain |$F_R$| by removing the existential quantifiers. Note that the computation of the look-up table takes constant time, since |$S$| is finite and not dependent on any input. We are now ready to transform |${\textit{KB}}$| into an appropriate |${S}^{\prime}$|-formula by applying the following replacement procedure as long as applicable. – Let |$C_R = R(x_1, \dots , x_n)$| be an |${S}$|-constraint (now the |$x_i$|’s are not necessarily |$n$| distinct variables). Replace |$C_R$| by its corresponding |${S}^{\prime}$|-formula |$F_R(x_1, \dots , x_n, y_1, \dots , y_m)$|⁠, where the variables |$y_1, \dots , y_m$| are fresh variables that are unique to |$C_R$| (they will not be used for any other constraint replacement). This transformation procedure introduces additional variables. We show that their total number is polynomially bounded. Denote by |$m_R$| the number of |$y_i$|’s added while replacing |$C_R$| (denoted |$m$| in the above procedure). One observes that the total number of additional variables is bounded by the number of original constraints times the maximum of all |$m_R$|⁠. Since |$m_R$| is only dependent on |$R$|⁠, it is constant. Since |$S$| is finite, the maximum of all |$m_R$| is constant. We conclude that the transformation can be achieved in polynomial time. Furthermore, observe that the so obtained abduction instance has exactly the same solutions as the original instance. The last lemma in this section takes care of the essentially positive cases. The proof idea is to remove the equality clauses maintaining the size counts and the satisfiability property. Lemma 3.4 Let |${S}, {S}^{\prime}$| be two constraint languages such that |${S}^{\prime}$| is essentially positive. Let |$\textrm{ABD}_* \in \{\textrm{ABD}, \textrm{ABD}_=, \textrm{ABD}_\leq\}$|⁠. If |${S} \subseteq \left \langle{S}^{\prime} \right \rangle $|⁠, then |$\textrm{ABD}_*(S) \leq^{\textbf{P}}_m \textrm{ABD}_*(S^\prime)$|⁠. Proof. The general case is due to Nordh and Zanuttini [31, Lemma 22]. The result for ‘|$\leq $|’ is because of the following. Removing equality clauses and deleting the duplicating occurrences of variables only decrease the size of |$H$|⁠. Now we proceed with proving the case for ‘=’. We show that for any |$\textrm{ABD}_=(S \cup \{=\})$| instance |${(V, H, M, {\textit{KB}},s)}{}$|⁠, there is an |$\textrm{ABD}_=(S)$|-instance |${(V_1, H_1, M_1, {\textit{KB}}_1,s)}{}$| such that the former has an explanation if and only if the later has one. The proof uses the fact that the only negative clauses in |${\textit{KB}}$| are of size |$1$|⁠. Since the existence of a solution is invariant under the equality clauses, we only need to assure that the size of a solution is also preserved. For each clause |$x_i= x_j \in{\textit{KB}}$|⁠, we do the following: If at most one of the |$x_i,x_j$| appears in |$H$|⁠, remove the clause |$x_i = x_j$| from |${\textit{KB}}$|⁠, replace |$x_j$| by |$x_i$| everywhere in |${\textit{KB}} \cup H\cup V \cup M$| (and delete |$x_j$|⁠). If both |$x_i, x_j$| are from |$H$|⁠. Then, – if |$\neg x_i$| (resp., |$\neg x_j$|⁠) appears in |${\textit{KB}}$|⁠, we add |$\neg x_j$| (resp., |$\neg x_i$|⁠) to |${\textit{KB}}$| and remove the clause |$x_i = x_j$| from |${\textit{KB}}$|⁠. otherwise |$\neg x_i,\neg x_j\notin{\textit{KB}}$| and then simply remove the clause |$x_i = x_j$| from |${\textit{KB}}$| and do not remove any variable. The problem caused by equality clauses is the following. If we remove a variable that is also a hypothesis, then removing this from |$H$|⁠, owing to some equality constraint, may not preserve the size of solutions. Furthermore, this problem occurs only when an equality clause contains both variables from |$H$| (case 2.), since otherwise the size of |$H$| is not changed (case 1.). We prove the following correspondence between the solutions of the two instances. Claim. A subset |$E\subseteq H$| is an explanation for |$\textrm{ABD}_=(S \cup \{=\})$| if and only if |$E$| is an explanation for |$\textrm{ABD}_=(S)$|⁠. Proof of Claim. ‘|$\!\!\!\!\implies\!\!\!\!$|’: Let |${(V,H,M,{\textit{KB}},s)}$| be an instance of |$\textrm{ABD}_=(S \cup \{=\})$| and let |${(V_1,H,M_1,{\textit{KB}}_1,s)}$| be the corresponding instance of |$\textrm{ABD}_=(S \cup \{=\})$|⁠, where |${\textit{KB}}_1$| is obtained from |${\textit{KB}}$| by applying steps 1 and 2 above. If |$E\land{\textit{KB}}$| is consistent, we prove that |$E\land{\textit{KB}}_1$| is also consistent. Note that |${\textit{KB}}_1 \subseteq{\textit{KB}}$|⁠, except if |$\neg x_j\in{\textit{KB}}_1$| for some |$x_j$|⁠. This implies that |$x_j\not \in E$| due to the reason that |$x_i = x_j$| and |$\neg x_i \in{\textit{KB}}$|⁠. Finally, |$M_1\subseteq M$| and |$E$| is an explanation for |$M$| implies |$E$| is also an explanation for |$M_1$|⁠. ‘|$\!\!\impliedby\! $|’: Suppose that |$E \land{\textit{KB}}_1$| is consistent and let |$\theta $| be a satisfying assignment. We consider each equality constraint separately and prove that |$E \land{\textit{KB}}$| is consistent. In the first case, for each |$x_i, x_j$| such that at most one (say |$x_i$|⁠) appears in |$H$|⁠. If |$x_i \in E \subseteq H$| then |$\neg x_j \not \in{\textit{KB}}_1$| since this would imply that |$\neg x_i \in{\textit{KB}}_1$| and |$x_i \not \in E$|⁠. Consequently, |$E \land{\textit{KB}}_1 \land (x_i=x_j)$| is consistent (by extending |$\theta $| to |$\theta (x_i)= 1 =\theta (x_j)$|⁠). On the other hand, if |$\neg x_j \in{\textit{KB}}_1$| then |$\neg x_i \in{\textit{KB}}_1$| and |$x_i \not \in E$|⁠. As a result, |$\theta $| extended to |$\theta (x_i)= 0 =\theta (x_j)$| is a satisfying assignment. In the second case, if both |$x_i, x_j \in H$| then we also have two sub-cases based on whether |$\neg x_i \in{\textit{KB}}_1$| or not. If |$\neg x_i \in{\textit{KB}}_1$| then due to case 2, we have |$\neg x_j\in{\textit{KB}}_1$| and this implies |$x_i\not \in E$|⁠. As a consequence, |$E \land{\textit{KB}}_1 \land (x_i=x_j)$| is consistent by extending |$\theta $| to |$\theta (x_i)= 0 =\theta (x_j)$|⁠. In the sub-case when both |$x_i, x_j \in H$| and |$\neg x_j \not \in{\textit{KB}}_1$| then mapping |$\theta (x_i)=1 = \theta (x_j)$| satisfies |$E\land{\textit{KB}}_1 \land (x_i= x_j)$|⁠. This is because all the non-unit clauses in |${\textit{KB}}_1$| are positive. Since this is true for all the equality clauses it shows that |$E\land{\textit{KB}}$| is consistent. For entailment, let |$m_i, m_j \in M$||$(m_i = m_j) \in{\textit{KB}}$| and |$m_j \not \in M_1$|⁠, i.e. |$M_1\subsetneq M$|⁠. Since |$E\land{\textit{KB}}_1$| is consistent and entails |$M_1$|⁠, we have |$E\land{\textit{KB}}_1 \land (m_i=m_j)$| is also consistent (due to arguments for consistency) and entails |$M_1 \cup \{m_j\}$|⁠. This completes the proof in this direction and settles the claim. Finally, the above reduction can be computed in polynomial time because both steps are applied once for each equality clause and each step takes polynomial time. This shows the desired reduction between |$\textrm{ABD}_=(S \cup \{=\})$| and |$\textrm{ABD}_=(S)$|⁠. REMARK 3.5 Notice that Lemmas 3.3 and 3.4 are stated with respect to the classical and unparameterized decision problems. However, these reductions can be generalized to |$\leq ^{\textbf{FPT}}$|-reductions whenever the parameters are bound as required by Def. 2.2. That is, in our case, for any parameterization |$k\in{\{\, |H|,|E|,|M| \,\}} $| the reductions are valid. Even more, the values of the parameters stay the same as in the reduction the sizes of |$H$|⁠, |$E$| and |$M$| remain unchanged. REMARK 3.6 It is rather cumbersome to mention the base independence results in almost every single proof. As a result, we omit this reference and show the results only for concrete bases, thereby, implicitly using the above lemmas. In cases where we deal with essentially negative constraint languages, we do not have a general base independence result but direct constructions showing membership and hardness in our cases for all bases (e.g. Lemmas 3.13 and 3.23). 3.2 General complexity results In this section, we start with general observations and reductions between the defined problems. Then we prove some immediate (parameterized) complexity results. We provide two results that help us to consider fewer cases to solve. Lemma 3.7 For every constraint language |$S$| we have |$\textrm{ABD}(S) \leq^{\textbf{P}}_m \textrm{ABD}_\leq(S)$|⁠. Proof. Clearly, |${(V,H,M,KB)} \in \textrm{ABD}(S) \Leftrightarrow{(V,H,M,KB,s)} \in \textrm{ABD}_\leq(S)$|⁠, where |$s=|H|$|⁠. That is, there is an explanation for an abduction instance if and only if there is one with size at most that of the hypotheses set. Lemma 3.8 |$\textrm{ABD}_\leq(S) = \textrm{ABD}_=(S)$| for any |${S}$| such that |${\textsf{IBF}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. Proof. ‘|$\subseteq $|’: We claim that every positive instance |${(V,H,M,{\textit{KB}},s)}\in \textrm{ABD}_\leq(S)$| has a solution |$E$| of size exactly |$s$|⁠. Given a solution of size |$\leq s$| then a solution of size |$=s$| can be constructed from it (in even polynomial time w.r.t. |$|H|$|⁠) by adding one element |$h$| at a time from |$H$| to |$E$| and checking that |$\neg h \not \in{\textit{KB}}$|⁠. ‘|$\supseteq $|’: Every solution of size exactly |$s$| is a solution of size |$\le s$|⁠. Intractable cases It turns out that for |$0$|-valid, |$1$|-valid and complementive languages, all three problems remain hard under any parameterization except the case |$|V|$|⁠. Theoem 3.9 The problems |$\textrm{ABD}(S, k)$|⁠, |$\textrm{ABD}_\leq(S, k)$|⁠, |$\textrm{ABD}_=(S, k)$| are |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$| and |$k\in{\{\, |H|,|E|, |M| \,\}} $|⁠, |${\textbf{para-}}{\textbf{DP}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$| and |$k\in \{|H|,|E|\}$|⁠. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |$k=|M|$| for |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠. Proof. (1.) We prove the case for |${\textsf{IN}}$| regarding all three parameters simultaneously. Notice that |${\textrm{IMP}}({\textsf{II}}_1)$| is |${\textbf{co}}{\textbf{NP}} $|-hard [31, Thm. 34] even if the right side contains only a single variable. We describe in the following a modified proof from [31, Prop. 48]. Since |$\left \langle{\textsf{IN}} \cup{\{\, T \,\}} \right \rangle = {\textsf{II}}_1$| (define |$T(x) \equiv x$|⁠) we have that |${\textrm{IMP}}({\textsf{IN}} \cup{\{\, T \,\}} )$| is |${\textbf{co}}{\textbf{NP}} \text{-hard} $|⁠, even if the right side contains only a single variable. We reduce |${\textrm{IMP}}({\textsf{IN}} \cup{\{\, T \,\}} )$| to our abduction problems with |$|H|=1$|⁠, |$|M|=1$| and |$|E|=1$|⁠. Let |$({\textit{KB}}_T, q)$| be an instance of |${\textrm{IMP}}({\textsf{IN}} \cup{\{\, T \,\}} )$|⁠, where |${\textit{KB}}_T = {\textit{KB}} \land \bigwedge _{x\in V_T} T(x)$| with |${\textit{KB}}$| being an |${\textsf{IN}}$|-formula. We map |$({\textit{KB}}_T, q)$| to |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime})}$|⁠, where |$V = \textrm{var}({\textit{KB}}) \cup \{h\}$|⁠, |$h$| is a fresh variable and |${\textit{KB}}^{\prime}$| is obtained from |${\textit{KB}}$| by replacing any variable from |$V_T$| by |$h$|⁠. Note that |${\textit{KB}}_T \equiv{\textit{KB}}^{\prime} \land h$|⁠. Since |${\textit{KB}}$| and |${\textit{KB}}^{\prime}$| are 1-valid, clearly, |${\textit{KB}}^{\prime} \land h$| is always satisfiable and there exists an explanation iff |${\textit{KB}}^{\prime} \land h \models q$|⁠, iff |${\textit{KB}}_T \models q$|⁠. Furthermore, observe that |${\textit{KB}}_T\models q$| if and only if |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime}, |H|)} \in \textrm{ABD}\textsf{IN},{|H|}$| if and only if |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime},1,|H|)} \in \textrm{ABD}_\leq\textsf{IN},{|H|}$| if and only if |${(V, \{h\}, \{q\}, {\textit{KB}}^{\prime},1,|H|)} \in \textrm{ABD}_=\textsf{IN},{|H|}$|⁠. The latter is true also when replacing |$|H|$| by |$|E|$| or |$|M|$|⁠. This proves the claimed |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hardnesses. (2.) From Fellows et al. [21, Prop. 4] we know that all three problems for |${\textsf{BR}}$| are |${\textbf{DP}} $|-complete for |$|H|=0$| even if |$|M|=1$|⁠. We argue that the hardness can be extended to |${\textsf{IN}}_2$|⁠. Note that |$\left \langle{\textsf{IN}}_2\cup \{F\} \right \rangle ={\textsf{BR}}$| where |$F(x) \equiv \neg x$|⁠. Creignou & Zanuttini [16] prove that |$\textrm{ABD}(S \cup \{F\}) \leq^{\textrm{P}}_m \textrm{ABD}(S \cup \{\texttt{SymOR}_{2,1}\})$| where |$\texttt{SymOR}_{2,1}(x,y,z)= ((x\rightarrow y) \land T(z)) \lor ((y\rightarrow x) \land F(z))$|⁠. Moreover, they also prove that |$\texttt{SymOR}_{2,1} \in \left \langle{S} \right \rangle $| such that |${\textsf{IN}}_2\subseteq \left \langle{S} \right \rangle $| [16, Lem. 21/27]. Finally, having |$|M|=1$| allows us to use their proof and, as a consequence, |$ \textrm{ABD}(\textrm{BR}) \leq ^{\textbf{P}}_m \textrm{ABD}(S)$| such that |${\textsf{IN}}_2 \subseteq \left \langle{S} \right \rangle $|⁠. This gives the desired lower bound for |${\textsf{IN}}_2$|⁠. Regarding |${\textsf{II}}_0$|⁠, the proof follows by a similar argument using the observations that |$\left \langle{\textsf{II}}_0 \cup{\{\, T \,\}} \right \rangle = {\textsf{BR}}$| and |$\texttt{OR}_{2,1} \in \left \langle{S} \right \rangle $| such that |${\textsf{II}}_0\subseteq \left \langle{S} \right \rangle $| where |$\texttt{OR}_{2,1}(x,y) = x\rightarrow y$| [16, Lem. 19/27]. (3.) Nordh and Zanuttini [31, Prop. 46/47] prove |${\boldsymbol{\varSigma _2^P}} $|-hardness for both |${\textsf{IN}}_2$| as well as |${\textsf{II}}_0$| with positive literal manifestations. This implies that the |$1$|-slice of each of |$\textrm{ABD}(\textsf{IN}_2,{|M|})$| and |$\textrm{ABD}(\textsf{II}_0,{|M|})$| is |${\boldsymbol{\varSigma _2^P}} $|-hard, which gives the desired result. For |$\textrm{ABD}_\leq(S, |M|)$| and |$\textrm{ABD}_=(S, |M|)$|⁠, the results follow from Lemma 3.7. Fixed-parameter tractable cases The following corollary is immediate because the classical questions corresponding to these cases are in |${\textbf{P}} $| due to Nordh and Zanuttini [31]. Corollary 3.10 The problem |$\textrm{ABD}(S, k)$| is |${\textbf{FPT}} $| for any parameterization |$k$| and |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}$| with |${\textsf{C}} \in{\{\, {\textsf{IV}}_2, {\textsf{ID}}_1, {\textsf{IE}}_1, {\textsf{IS}_{12}} \,\}} $|⁠. The next result is already due to Fellows et al. [21, Prop. 13]. Corollary 3.11 The problems |$\textrm{ABD}(S, |V|)$|⁠, |$\textrm{ABD}_\leq(S, |V|)$|⁠, |$\textrm{ABD}_=(S, |V|)$| are all |${\textbf{FPT}} $| for all Boolean constraint languages |${S}$|⁠. Now, we prove |${\textbf{P}} $|-membership for some cases of the classical problems and start with the essentially positive cases. The proof idea is to start with unit propagation. The positive clauses do not explain anything and one just only checks whether the elements of |$M$| appear either in |${\textit{KB}}$| or |$H$|⁠. Then, we need to adjust the size accordingly. Lemma 3.12 The classical problems |$\textrm{ABD}_=(S)$| and |$\textrm{ABD}_\leq(S)$| are in P for |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{02}}$|⁠. Proof. We prove the claim for |$\textrm{ABD}_=(S)$|⁠, whereas the result for |$\textrm{ABD}_\leq(S)$| is due to Lemma 3.8. Let |${(V,H,M,{\textit{KB}},s)}\in \textrm{ABD}_=(S)$| be an instance where |${S}\subseteq{\textsf{IS}_{02}}$|⁠. Denote by |$ H^{\prime},M^{\prime},{\textit{KB}}^{\prime}$| the result of applying the unit propagation on those literals |$y$| such that |$y \in \textrm{Lit}({\textit{KB}}) \backslash (H^{+} \cup M^{-})$|⁠. Recall that for a set |$Y$| of literals, |$Y^{+}$| (resp., |$Y^{-}$|⁠) denotes the set of positive (negative) literals formed upon |$Y$|⁠. In unit propagation, for a unit clause |$u$|⁠, any clause containing |$u$| can be deleted and delete in any clause |$\sim \!u$|⁠, where |$\sim \!u=x$| if |$u=\lnot x$| is a negative literal and |$\sim \!u=\lnot x$| if |$u=x$| is a positive literal. Note that literals |$y \in H^{+}\cup M^{-}$| (i.e. |$y\in H$| or |$y=\neg m$| with |$ m \in M$|⁠) are excluded from this rule as mentioned above. The reason for this choice is as follows. If |$\neg m \in{\textit{KB}}$| for some |$m\in M$| then removing |$m$| from |${\textit{KB}}\cup M$| transforms a ‘no solution’—to a ‘yes solution’—instance. Similarly, removing an |$h\in H$| from |${\textit{KB}}\cup H$| may decrease the solution size of the instance. Finally, the positive literal |$m\in M$| may or may not be processed. However, it is important to consider |$h\in H^{-}$| since this helps in invalidating the clauses of length |$\geq 2$|⁠. Let |$P$| and |$N$| be the positive, respectively negative unit clauses of |${\textit{KB}}^{\prime}$| over |$\left \langle{S} \right \rangle $|⁠. Note that if |$N\not =\emptyset $| then there can be no explanation for |$M$|⁠. This is due to the fact that only negative unprocessed literals are over |$M$| implying that |${\textit{KB}}$| is inconsistent with |$M$|⁠. Because of this, we have |$N=\emptyset $|⁠. Moreover, the positive clauses of length |$\geq 2$| in |${\textit{KB}}^{\prime}$| do not explain anything as a variable cannot be enforced |$0$|⁠. Therefore, a positive literal |$x$| cannot explain anything more than |$x$| itself. This implies that there is an explanation for |$M$| if and only if |$M^{\prime} \subseteq H^{\prime} \cup P$|⁠. Now, the set |$M^{\prime}\setminus P$| denotes those |$m\in M$| that are not already explained by |${\textit{KB}}$| and must be explained by |$H^{\prime}$|⁠. As a consequence, there exists an explanation for |$\textrm{ABD}_\leq(S)$| if and only if |$M^{\prime}\setminus P \subseteq H^{\prime}$| and |$|M^{\prime}\setminus P| \leq s$|⁠. The consistency is already assured by the fact that |$N=\emptyset $|⁠. Finally, to determine whether there is an explanation |$E \subseteq H$| of size |$s$|⁠, it suffices to check additionally whether |$|H^{\prime}| \geq s$|⁠. This argument ensures whether we can artificially increase the solution size, since, in that case an |$E \subseteq H^{\prime}$| with above conditions constitutes an explanation for the problem . If this is not true, then no explanation of size |$s$| exists. The unit propagation and the size comparisons can be done in polynomial time, which proves the claim. The following lemma proves that essentially negative languages for |$\textrm{ABD}_\leq$| also remain tractable. Lemma 3.13 The classical problem |$\textrm{ABD}_\leq(S)$| is in P if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠. Proof. First, we prove the result with respect to |$\left \langle S \right \rangle _{\neq }\subseteq{\textsf{IS}_{12}}$|⁠. Let |$P$| denote the set of positive unit clauses from |${\textit{KB}}$| and denote |$E_{MP} = M \setminus P$|⁠. Now, we have the following two observations. Observation 1 There exists an explanation iff |$E_{MP} \subseteq H$| and |$M$| is consistent with |${\textit{KB}}$|⁠. That is, what is not yet explained by |$P$| must be explainable directly by |$H$| because negative clauses cannot contribute to explaining anything, they can only contribute to ‘rule out’ certain subsets of |$H$| as possible explanations. Observation 2 If there exists an explanation, then any explanation contains |$E_{MP}$|⁠. As a result, |$E_{MP}$| represents a cardinality-minimal and a subset-minimal explanation. We conclude that there exists an explanation |$E$| with |$|E| \leq s$| iff |$E_{MP}$| constitutes an explanation and |$|E_{MP}| \leq s$|⁠. Now, we proceed with base independence for this case. Claim. |$\textrm{ABD}_\leq(S \cup \{=\}) \leq^{\textbf{P}}_m \textrm{ABD}_\leq(S)$| for |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠. Proof of Claim. The reduction gets rid of the equality clauses by removing them and deleting the duplicating occurrences of variables. This decreases only the size of |$H$| and might also the size of an explanation |$E$|⁠. Notice that |$x=y\in{\textit{KB}}$| does not enforce both |$x$| and |$y$| into |$E$|⁠. This completes the proof to the lemma. Finally, the |$2$|-affine cases are also tractable as we prove in the following lemma. The idea is, similar to Creignou et al. [10, Prop. 1], to change the representation of the knowledge base. Lemma 3.14 The classical problems |$\textrm{ABD}_=(S)$| and |$\textrm{ABD}_\leq(S)$| are in P if |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_1$|⁠. Proof. Analogously to Creignou et al. [10, Prop. 1], we change the representation of the |${\textit{KB}}$|⁠. Without loss of generality, suppose |${\textit{KB}}$| is satisfiable and contains no unit clauses since unit clauses can be dealt with in a straightforward way. Each clause expresses either equality or inequality between two variables. With the transitivity of the equality relation and the fact that (in the Boolean case) |$a \neq b \neq c$| implies |$a = c$|⁠, we can identify equivalence classes of variables such that each two classes are either independent or they must have contrary truth values. We call a pair of dependent equivalence classes |$(X, Y)$| a cluster (⁠|$X$| and |$Y$| must take contrary truth values). Denote by |$X_1, \dots , X_p$| the equivalence classes that contain variables from |$M$| such that |$X_i \cap M \neq \emptyset $|⁠. Denote by |$Y_1, \dots , Y_p$| the equivalence classes such that for each |$i$| the pair |$(X_i, Y_i)$| represents a cluster. We make the following stepwise observations. There is an explanation iff |$\forall i: H \cap X_i \neq \emptyset $|⁠. The size of a minimal explanation (⁠|$E_{min}$|⁠) is |$p$|⁠, it is constructed by taking exactly one representative from each |$X_i$|⁠. There exists an explanation of size |$\leq s$| iff |$p \leq s$|⁠. An explanation of maximal size (⁠|$E_{max}$|⁠) can be constructed as follows: (a) |$E:= \emptyset $|⁠, (b) for each |$i$| add to |$E$| all variables from |$X_i \cap H$|⁠, (c) for each cluster |$(X,Y) \notin \{(X_i, Y_i) \mid 1 \leq i \leq p\}$|⁠: if |$|X\cap H| \geq |Y\cap H|$|⁠: add to |$E$| the set |$X\cap H$|⁠, else: add to |$E$| the set |$Y\cap H$|⁠. Any explanation size between |$|E_{min}|$| and |$|E_{max}|$| can be constructed. There is an explanation of size |$=s$| iff |$|E_{min}| \leq s \leq |E_{max}|$|⁠. This completes the proof. Lemmas 3.12–3.14 imply the following corollary. Corollary 3.15 The following problems are FPT for any |$k\in{\{\, |H|,|E|, |M| \,\}} $|⁠. |$\textrm{ABD}_=(S, k)$| if |$\left \langle{S} \right \rangle \subseteq C$| for |${\textsf{C}} \in{\{\, {\textsf{IS}_{02}},{\textsf{ID}}_1 \,\}} $|⁠, |$\textrm{ABD}_\leq(S, k)$| if |$\left \langle{S} \right \rangle \subseteq C$| for |${\textsf{C}} \in{\{\, {\textsf{IS}_{02}},{\textsf{ID}}_1,{\textsf{IS}_{12}} \,\}} $|⁠. Now we move towards the parameter specific results for each problem. 3.3 Parameter ‘number of hypotheses’ |$|H|$| For this parameter, it turns out that the only intractable cases are those pointed out in Lemma 3.9. Theoem 3.16 |$\textrm{ABD}(S, |H|)$|⁠, |$\textrm{ABD}_\leq(S, |H|)$| and |$\textrm{ABD}_=(S, |H|)$| are |${\textbf{para-}}{\textbf{DP}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$|⁠, |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in{\{\, {\textsf{IE}}_2, {\textsf{IV}}_2, {\textsf{ID}}_2,{\textsf{IL}}_2 \,\}} $|⁠. Proof. (1.+2.) Follows from Lemma 3.9. (3.) Recall that |${\textrm{SAT}}({S})\ \textrm{and}\ {\textrm{IMP}}({S})$| are both in |${\textbf{P}} $| for every |${S}$| in the question (Prop. 2.7). By |$|H|\ge |E|$|⁠, we have that |$\binom{|H|}{|E|}=|H|^{|E|}\in O(k^k)$|⁠, where |$k=|H|$|⁠. Consequently, we brute-force the candidates for |$E$| and verify them in polynomial time. This yields |${\textbf{FPT}} $| membership. 3.4 Parameter ‘number of explanations’ |$|E|$| In this subsection, we consider the solution size as a parameter. Notice that, because of the parameter |$|E|$|⁠, the parameterized version of the problem |$\textrm{ABD}$| is not meaningful anymore. As a result, we only consider the size limited variants |$\textrm{ABD}_=$| and |$\textrm{ABD}_\leq $|⁠. The following theorem provides a classification of both problems into six different complexity degrees. Theoem 3.17 The problems |$\textrm{ABD}_\leq(S, |E|)$| and |$\textrm{ABD}_=(S, |E|)$| are |${\textbf{para-}}{\textbf{DP}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$| |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, |${\textbf{W}\textbf{P}}$|-complete if |${\textsf{IE}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IE}}_2$|⁠, |$\textbf{W}[2]$|-complete if |${\textsf{IM}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{C}}$| for |${\textsf{C}} \in \{{\textsf{ID}}_2,{\textsf{IS}^{\ell }_{10}}, {\textsf{IV}}_2\}$|⁠, |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_1$| or |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{02}}$|⁠, Moreover, if |${\textsf{IS}^{2}_{1}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}} $|⁠, then |$\textrm{ABD}_\leq(S, |E|) \in \textbf{FPT}$| and |$\textrm{ABD}_=(S, |E|)$| is |$\textbf{W}[1]$|-complete. Proof. (1.+2.) Follows from Lemma 3.9. (3.) The upper bound for |${\textsf{IE}}_2$| follows from the fact that |${\textrm{SAT}}({\textsf{IE}}_2)$| and |${\textrm{IMP}}({\textsf{IE}}_2)$| are in |${\textbf{P}} $| (cf. Prop. 2.7). Guessing |$E$| takes |$k \cdot \log n$| non-deterministic steps and verification can be done in polynomial time. For the lower bound, we argue that the proof from [21, Cor. 9] for definite Horn theories (⁠|${\textsf{IE}}_1$|⁠) can be extended. The only types of clauses used are |$x \land y \rightarrow z$| and |$x \rightarrow y$|⁠, which are both in |${\textsf{IE}}$| and consequently expressible by |${S}$| as |${\textsf{IE}} \subseteq \left \langle{S} \right \rangle $|⁠. Both membership and hardness arguments are valid for |$\textrm{ABD}_\leq(S, |E|)$| as well (the problem in [21, Cor. 9] used for hardness is Monotone Circuit SAT, which is monotone). (4.) The completeness for |$\textrm{ABD}_=(S, |E|)$| such that |$\left \langle{S} \right \rangle ={\textsf{IM}}$| follows from Lemma 3.18. Lemma 3.19 strengthens the result by showing -membership of the problem |$\textrm{ABD}_=(S, |E|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. The question |$\textrm{ABD}_\leq(S, |E|)$| for the above two cases follow from the monotone argument of Lemma 3.8. Moreover, for |${\textsf{ID}}_2$|⁠, the result follows from [21, Thm. 21]. The membership for |$\textrm{ABD}_=(S, |E|)$| and |$\textrm{ABD}_\leq(S, |E|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$| is due to Lemmas 3.20 and 3.21, respectively. The hardness for both cases follows from Lemma 3.18. (5.) Follows from Corollary 3.15. Finally, the |${\textbf{FPT}} $| membership for |$\textrm{ABD}_\leq\textsf{IS}_{12}, {|E|}$| is shown in Corollary 3.15. The |$\textbf{W}[1]$|-hardness for |$\textrm{ABD}_=(S, |E|)$| with |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$| follows from Lemma 3.22 where we prove |$\textbf{W}[1]$|-hardness for the languages |${S}$|⁠, such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠. The |$\textbf{W}[1]$| membership for |$\textrm{ABD}_=(S, |E|)$| with |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠, this means also the arbitrary bases, is due to Lemma 3.23 This concludes all the cases in Theorem 3.17. 3.4.1 Intermediate Lemmas Note that the difficult part of the abduction problem for |${S}$| such that |${\textsf{IM}} \subseteq \left \langle{S} \right \rangle $| is the case when a solution of size larger than |$k$| is found. This solution must be reduced to one of size |$\leq k$| (resp. |$=k$|⁠). First we prove the completeness for the language |${S}$| such that |$\left \langle{S} \right \rangle ={\textsf{IM}}$|⁠. Later we extend the membership result to the languages in |$ {\textsf{IV}}_2$| and |${\textsf{IS}^{l}_{10}}$|⁠. Lemma 3.18 |$\textrm{ABD}_=(S, |E|)$| is |$\textbf{W}[2]$|-complete if |$\left \langle{S} \right \rangle ={\textsf{IM}}$|⁠. Proof. For membership we prove that |$\textrm{ABD}_=(\textsf{IM}, |E|) \leq^{\textbf{FPT}} {\textrm{p-WSAT}(\varGamma _{2,1})}$|⁠. The latter is known to be |$\textbf{W}[2]$|-complete (Proposition 2.4). Let |${(V,H,M,{\textit{KB}},k)}$| be an instance of |$\textrm{ABD}_=(\textsf{IM}, |E|)$|⁠, where the solution size is the parameter (i.e. |$s=k$|⁠). Specifically, let |${\textit{KB}}=\bigwedge \limits _{i\leq r}(x_i \rightarrow y_i)$| and |$M= m_1\land \ldots \land m_{|M|}$|⁠. Note that, in order to explain a single |$m_i\in M$|⁠, a single |$h\in H$| suffices. As a result, for each |$m_i\in M$| we associate a set |$H_i \subseteq H$| of hypotheses that explains |$m_i$|⁠. This implies that every element (singleton subset) of |$H_i$| explains |$m_i$|⁠. Now, it is enough to check that at least one such |$h\in H_i$| can be selected for each |$m_i$|⁠. For this we map |${(V,H,M,{\textit{KB}},k)}$| to |$(\phi ,k)$| where |$\phi = \bigwedge \limits _{i\leq |M|} \bigvee \limits _{x\in H_i} x$|⁠. Then our claim is that |${(V,H,M,{\textit{KB}},k)}$| has an explanation |$E$| if and only if |$\phi $| has a satisfying assignment of size |$k$|⁠. Clearly, there is a |$1-1$|-correspondence between solutions |$E$| of |${(V,H,M,{\textit{KB}},k)}$| and satisfying assignments |$\theta $| with weight |$k$| for |$\phi $|⁠. That is, |$\theta (x)=1 \iff x\in E$|⁠. For hardness, we reduce from |${\textrm{p-WSAT}(\varGamma ^+_{2,1})}$|⁠, which is |$\textbf{W}[2]$|-complete by Proposition 2.4. Given |$\bigwedge \limits _{i\in q}\bigvee \limits _{j\in r}(X_{ij}^+) = \bigwedge \limits _{i\in q}(X_{i_1}\lor \ldots \lor X_{i_r})$|⁠, where |$\textrm{var}(\alpha )={\{\, X_{ij}\mid i\in q, j\in r \,\}} $|⁠, we let |${\textit{KB}}= \bigwedge \limits _{i\in q}\bigwedge \limits _{x\in h_i}(x\rightarrow h_i$|⁠, |$H=\textrm{var}(\alpha )$|⁠, |$M= \bigwedge \limits _{i\in q}h_i $| and |$V=H\cup M$|⁠. Then for a subset |$E\subseteq H$| we have that, |$E$| is an explanation for |$\textrm{ABD}_=(\textsf{IM}, |E|) \iff \theta \models \phi $| where |$\theta (x)=1 \iff x\in E$|⁠. Now we show that with a little modification, the same reduction (as for the membership of |${\textsf{IM}}$|⁠) can be used to prove |$\textbf{W}[2]$|-membership for |${\textsf{IV}}_2$|⁠. Lemma 3.19 |$\textrm{ABD}_=(S, |E|)$| is in |$\textbf{W}[2]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. Proof. In the IM-case, we dealt only with clauses of type |$x \rightarrow y$|⁠. We refer to such classes as type-|$0$| clauses. In |${\textsf{IV}}_2$| we have additional clauses of the following types: Unit clauses: both positive and negative. |$x$|⁠, |$\neg x$| Positive clauses of size two or greater: |$(x_1 \lor \dots \lor x_n)$|⁠, |$n\geq 2$| Clauses with exactly one negative literal of size |$3$| or greater: |$(\neg x_0 \lor x_1 \lor \dots \lor x_n)$|⁠, |$n\geq 2$| We can we eliminate the type-|$1$| clauses by unit propagation and obtain thereby a satisfiability equivalent formula. Note that this transformation process can generate additional clauses of type-|$0$|⁠, type-|$2$| or type-|$3$|⁠. As a consequence, we end up only with clauses of either type-|$0$|⁠, type-|$2$| or type-|$3$| and, particularly, no type-|$1$| clauses anymore. This transformation does not preserve all the satisfying assignments but those can be maintained by adding fixed values of the eliminated variables to the assignment. Now, we argue that by applying resolution on the variables in |${\textit{KB}} \setminus H$|⁠, we can ignore type-2 and type-3 clauses. Notice that we do not apply resolution to the variables in |$H$|⁠. Recall the idea behind the construction of Lemma 3.18, we want to come up with a formula that has a satisfying assignment if and only if our abduction instance has an explanation. A satisfying assignment that selects a variable |$x \in H$| (maps |$x$| to |$1$|⁠) forces all the variables |$y_1,\ldots , y_n$| such that |$(x\rightarrow y_i) \in{\textit{KB}}$| to be mapped |$1$| for |$i\leq n$|⁠. Furthermore, it also forces each |$z_{i,j}$| such that |$(y_i \rightarrow z_{i,j}) \in{\textit{KB}}$| to be mapped |$1$|⁠, and so on. This precisely captures the intuition that |$x$| (as a hypothesis) explains each |$y_i$| and |$z_{i,j}$|⁠. As a consequence, removing such variables from |$H$| (owing to resolution) in the case when those variables explain some manifestation would be problematic. Finally, we prove the claim that for |$\textrm{ABD}_=(\textsf{IV}_2, |E|)$| we can ignore the type-|$2$| and type-|$3$| clauses. Type-2 clauses are irrelevant since the satisfaction of such clauses does not force any particular variable to |$1$|⁠. In a type-3 clause (⁠|$C= \neg x_1 \lor x_2 \lor \dots \lor x_m$|⁠) the variable |$x_1$| forces a whole clause to be true (at least one of the remaining variables must be mapped to |$1$|⁠). Such clauses cannot be ignored right away because there might be further clauses of the form |$\neg x_j \lor m$| for each |$2\leq j \leq m $| with |$m\in M$| and an explanation to |$m$| might be lost (selecting |$x_1$| in the solution). However, after applying resolution we know that type-3 clauses only force one of the many positive variables to |$1$| and do not actually force a single variable to |$1$|⁠. As a result, this allows us to ignore type-3 clauses as well. Consequently, we are only left with type-0 clauses. This completes the proof by the same arguments as in the proof of Lemma 3.18. The question |$\textrm{ABD}_\leq(S, |E|)$| for the above two cases follows from the monotone argument of Lemma 3.8. Moving forward, the hardness for |${\textsf{IS}^{\ell }_{10}}$| is a consequence of the |$\textbf{W}[2]$|-hardness for IM. However, we strengthen this results for |$\textrm{ABD}_=$| to |$\textbf{W}[2]$|-completeness by showing the membership in |$\textbf{W}[2]$|⁠. Lemma 3.20 Let |$\ell \ge 2$|⁠, then |$\textrm{ABD}_=(S, |E|)$| is in |$\textbf{W}[2]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. We reduce our problem to |${\textrm{p-WSAT}(\varGamma _{2,1})}$|⁠, which is |$\textbf{W}[2]$|-complete (Prop. 2.4). Consider the reduction from Lemma 3.18 again, where we map |${(V,H,M,{\textit{KB}},s)}$| to |$(\phi ,k)$|⁠, where |$\phi = \bigwedge \limits _{i\leq m} \bigvee \limits _{x\in H_i} x$|⁠. The only difference from Lemma 3.18 is that in |${\textsf{IS}^{\ell }_{10}}$| there are additional constraints of the form |$(\neg x_1 \lor \ldots \lor \neg x_q)$| where |$q\leq \ell $|⁠. Now we have two cases. If all the additional constraints contain exclusively variables from H then we simply add these constraints to |$\phi $| and obtain a new formula |$\psi $|⁠. Since any satisfying assignment for |$\psi $| would satisfy these constraints as well as |$\phi $| and therefore is an explanation as required. Conversely, any explanation would yield a satisfying assignment for this new formula |$\psi $| since this explanation is consistent with |${\textit{KB}}$|⁠. Now suppose that constraints contain variables that are not from |$H$|⁠. We transform such constraints into their equivalents, which contain variables only from H. To achieve this we repeat the following procedure as long as applicable: Pick a variable |$u \not \in H$| occurring in a constraint |$C_u$|⁠. Compute the set of hypotheses |$H_u \subseteq H$| that explain |$u$| (analogously to Lemma 3.18). Let |$H_u = {\{\, h_1,\ldots , h_r \,\}} $|⁠. Now we replace the constraint |$C_u$| by |$r$| copies of itself and in each |$C_u^i$| we replace the variable |$u$| by |$h_i$|⁠. Note that this does not change the width of any clause. Finally, we add these clauses to |$\phi $| and obtain a new formula |$\psi $|⁠. Claim. The above construction preserves the correspondence between the solutions of |$\textrm{ABD}_=({\textsf{IS}^\ell_{10}}, |E|)$| and the satisfying assignments of |$\phi $| with weight |$k$|⁠. Moreover, it can be achieved in polynomial time. Proof of Claim. Note that the difference between Lemma 3.18 and this case is in the fact that a solution to |$\textrm{ABD}_=({\textsf{IS}^\ell_{10}}, |E|)$| must satisfy additional constraints as specified above. The problematic part is when some variables |$x_i,\ldots x_j$| are in |$H$| and some constraint over these variables appears in the |${\textit{KB}}$|⁠. The formula |$\psi $| must not allow such elements to be the part of solution since the constraints stop certain elements to appear together in a solution (being negative clauses). This proves the first claim in conjunction with the arguments in Lemma 3.18. Now we prove that this transformation works in polynomial time. The worst case is when a clause contains no variable from H. Furthermore assume that this clause is of maximum arity, say |$C= (\neg x_1,\ldots \neg x_q)$| where |$q\leq \ell $| and |$q$| is the maximum arity of constraint language in |${\textit{KB}}$|⁠. Each |$x_i$| can have the associated set |$H_{x_i}$| of maximum size |$n$| where |$n$| is input size. Hence each clause will be blown-up to at most |$n^q$| new constraints at the completion of the above procedure. As |$q$| is constant (only depends on the constraint language and not on the input), the factor |$n^q$| is polynomial. Since there are polynomial many constraints to check for this procedure, we conclude that the transformation takes only polynomial time. Eventually, similar arguments as in Lemma 3.18 for |$\psi $| complete the proof. We wish to point out that the reduction in Lemma 3.20 does not immediately settle the complexity for |$\textrm{ABD}_\leq(S, |E|)$| for |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. In the following lemma, we reduce |$\textrm{ABD}_\leq(S, |E|)$| to |${\textrm{Short-NTM-Halt}}$|⁠, the halting problem for non-deterministic multi-tape Turing machines. The input to the problem is |$(\mathbb M, k)$|⁠, where |$\mathbb M$| is a non-deterministic Turing machine, |$k$| is the parameter and the task is to decide whether |$\mathbb M$| accepts the empty string in at most |$k$| steps. This problem is |$\textbf{W}[2]$|-complete [22, Thm. 7.28]. The following reduction provides the |$\textbf{W}[2]$|-membership for |$\textrm{ABD}_\leq(S, |E|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠, the hardness follows from Lemma 3.18. Lemma 3.21 Let |$\ell \ge 2$|⁠, then |$\textrm{ABD}_\leq(S, |E|)$| is in |$\textbf{W}[2]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. The proof is in fact an extension of Lemma 3.20. Proceeding as before, we map |${(V,H,M,{\textit{KB}},k)}$| to |$(\psi ,k)$|⁠, where |$\psi $| is a collection of positive and negative clauses. Let |$U_i$| for |$i\leq P$| (resp., |$V_j, j\leq N$|⁠) denote the collection of positive (negative) clauses and |$L=P+N$|⁠. In the following, we reduce our problem to the problem |${\textrm{Short-NTM-Halt}}$| for multi-tape Turing machines. The reduction provides a multi-tape NTM |$\mathbb{M}$| such that |$\psi $| has a satisfying assignment of size at most |$k$| if and only if |$\mathbb{M}$| accepts the empty string in at most |$f(k)$| steps. |$\mathbb{M}$| has |$L+1$| tapes and there is one tape per each clause. The initial |$P$| tapes are dedicated to the positive clauses and the following |$N$| tapes to the negative ones. For convenience, we name the |$i$|th tape corresponding to a positive clause, as |$u_i$| (⁠|$i\leq P$|⁠) and |$j$|th tape corresponding to a negative clause, as |$v_j$| (⁠|$j\leq N$|⁠). The last tape is referred to as the tape |$L+1$|⁠. Intuitively, the computation of |$\mathbb{M}$| has the following four phases. Mark the length of each negative clause |$V_j$| on the tape |$v_j$|⁠, where |$j\leq N$|⁠. At the same time, write non-deterministically |$k$| elements |$x_1,\ldots ,x_k$| from |$V$| on the tape |$L+1$|⁠. Remove the duplicate variables from the tape |$L+1$|⁠. Read the tape |$L+1$|⁠. At the same time, for each tape |$w_r$|⁠, mark the cells if the elements being read appears in the respective clause |$W_r$| where |$r\leq L$|⁠. If at least one cell of each tape |$u_i$| (⁠|$i\leq P$|⁠) is marked and not every cell of the tape |$v_j$| (⁠|$j\leq N$|⁠) is marked, then accept. We first claim that the negative clauses of length |$\geq k$| can be ignored. The reason is that, for any assignment that maps |$k$| variables to |$1$|⁠, there are still literals in such clauses that are mapped to |$0$|⁠. Consequently, these clauses are trivially satisfied. This implies that the length of each tape corresponding to a negative clause is bounded by |$k$|⁠. For positive clauses, the length does not matter. This is because, our machine only needs to determine if at least one variable appearing in each positive clause is mapped to |$1$| in the assignment. Consequently, the machine only reads at most |$k$| cells on each of its tape. Our construction requires that the length of each |$V_j$| is hardcoded on the tape |$v_j$| for |$j\leq N$|⁠. This ensures that |$\mathbb{M}$| runs in parallel and does not need a state set of exponential size to ensure the correct computation. For each |$V_j$| of length |$l_j$|⁠, the length of the tape |$v_j$| is |$l_j+1$|⁠, where |$j\leq N$|⁠. This is achieved through having a collection of |$r+1$| states, where |$r = \textrm{max}\left |l_j \mid j \leq N\right |$| and |$r\leq k$|⁠. Moreover, even though |$\mathbb M$| can guess duplicate elements, it must work with the distinct collection of variables in the subsequent steps. That is, the multiple occurrences of any variable should be removed from the guessed assignment. A detailed but high-level description of |$\mathbb M$| is given below. – In the first |$k$| steps, the head of each tape |$v_j $| writes the symbol ‘*’ for |$l_j$|-cells and the symbol ‘#’ in the cell |$l_j+1$|⁠, where |$l_j$| is the length of |$V_j$| for |$j\leq N$|⁠. In the same steps, the head of the tape |$(L + 1)$| non-deterministically writes |$k$| elements |$x_1, \ldots , x_k$| from |$V$|⁠, into the first |$k$| cells. After |$k$| steps, the heads go back to the first cell. In the next (at most) |$k^2$| steps, the head on the tape |$L+1$| removes any duplicates. In the following |$h$| steps (where |$h\leq k$|⁠), the head of the tape |$(L + 1)$| reads the guessed elements (without duplicates) |$x_1, \ldots , x_h$| and at the same time, in |$r$|th of these steps, the head of each tape |$u_i$| determines whether |$U_i$| contains |$x_r$|⁠, for |$i\leq P $|⁠, while the head of each tape |$v_j$| determines if |$\neg x_r$| appears in |$V_j$| for |$j\leq N $|⁠. In the first case, it writes ‘|$yes$|’ in the cell, in the latter cases, it does not write. After writing in each case, the head moves right. If the variables do not appear in the corresponding clauses, the heads neither move nor print. After the previous |$h\leq k$| steps, the head on tapes |$u_i$| for |$i\leq P$| moves one step left while for other tapes, it stays in the same cell. If the head reads ‘|$yes$|’ in the |$u_i$|-tapes (⁠|$i\leq P$|⁠) and a ‘*’ in |$v_j$|-tapes (⁠|$j\leq N$|⁠), then |$\mathbb M$| accepts. Claim. |$\mathbb{M}$| can be constructed from |$\psi $| in FPT-time. Moreover, |$\mathbb M$| accepts the empty input in at most |$ k^2 + 3k+2$| steps if and only if |$\psi $| has a satisfying assignment of size at most |$k$|⁠. Proof of Claim. The number of tapes is |$L+1$| where |$L$| is the number of clauses in |$\psi $|⁠. The alphabet of |$\mathbb M$| constitutes |$V\cup \{yes,*,\#\}$| where |$V$| is the collection of variables in |$\psi $|⁠. The set of states has size |$k+O(1)$|⁠, |$k$| of which are required to ensure the first phase of the computation, i.e. guessing |$k$| elements and marking the tapes |$v_j$| for |$j\leq N$|⁠. Recall that the computation of |$\mathbb M$| completes in four phases. In the first phase, the head on the tape |$L+1$| moves right, writing |$x\in V$| (non-deterministically). Whereas, the head on the tape |$v_j$| (⁠|$j\leq N$|⁠) writes a ‘*’ for |$l_j$| many cells where |$l_j\leq k$| is the length of |$V_j$| and a ‘#’ in the last cell; moreover, the head stays in the last cell. When moving backwards, the head on the tape |$L+1$| can read any element |$x\in V$|⁠. However, the head on each |$v_j$|-tape (⁠|$j\leq N$|⁠) reads the symbol ‘#’ exactly once and the symbol ‘*’ in the remaining cells. The transitions for |$\mathbb M$| force every head to move one step left when reading ‘#’ and then the heads can only read the symbol ‘*’. This implies that the transition relation has the size |$O(k\cdot |\psi |^2)$| for the first phase. Finally, the transitions for ‘removing duplicates’, ‘comparing the variables’ and ‘the final check’ each has size |$O(k\cdot |\psi |)$|⁠. Therefore |$\mathbb M$| can be constructed from |$\psi $| in FPT-time. This proves the first part of the claim. For the second part, notice that the machine runs for |$2k + k^2 + k+2$| many steps. The first |$2k$| steps account for marking the length of each negative clause on the corresponding tapes and for guessing |$k$| elements on tape |$L+1$|⁠. In both cases the head of each tape should move back to read the first cell (the reason for |$2k$| steps). The following |$k^2$| steps are required to determine and remove the duplicate variables from the guessed list of variables. Lastly, at most |$k$| steps are required to compare the variables against each clause and the final two steps determine the accepting criteria for each tape. For the correctness, notice that the following three statements are equivalent. |$\mathbb M$| guesses |$k$| elements in such a way that the head of the tape |$u_i$| (⁠|$i\leq P$|⁠) reads ‘|$yes$|’, no head of the tape |$v_j$| (⁠|$j\leq N$|⁠) reads ‘#’ and the machine halts in the accepting state. The assignment (of weight at most |$k$|⁠) guessed by |$\mathbb M$| is such that there is at least one variable per each positive clause and for each negative clause, the assignment does not contain all of its variables. |$\psi $| has a satisfying assignment |$s$| of weight at most |$k$|⁠, |$s$| contains at least one variable from each positive clause and none of the negative clause contains all the variables appearing in |$s$|⁠. Consequently, if |$\psi $| has a satisfying assignment |$s$| of weight |$k$|⁠, then |$\mathbb M$| simply guesses this assignment and halts in the accepting state. Conversely, if |$\mathbb M$| accepts, then the guessed elements constitute a satisfying assignment for |$\psi $|⁠. This completes the proof to Lemma 3.21. Regarding the parameter |$|E|$|⁠, the only cases where |$\textrm{ABD}_\leq(S, |E|)$| and |$\textrm{ABD}_=(S, |E|)$| have different complexity is when |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}_{12}}$|⁠. The problem |$\textrm{ABD}_\leq(S, |E|)$| is FPT (Corollary 3.15). In the following lemmas, we prove |$\textbf{W}[1]$|-completeness for |$\textrm{ABD}_=(S, |E|)$|⁠. The |$\textbf{W}[1]$|-hardness for |$\textrm{ABD}_=(S, |E|)$| is proven for the languages |${S}$| such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠. Lemma 3.22 For any constraint language |${S}$| such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠, the problem |$\textrm{ABD}_=(S, |E|)$| is |$\textbf{W}[1]$|-hard. Proof. The problem IndependentSet is known to be |$\textbf{W}[1]$|-hard [17]. We reduce IndependentSet to |$\textrm{ABD}_=(S, |E|)$|⁠. Let |$((V, \tilde E), k)$| be an instance of |$p$|-IndependentSet and |$k$| the parameter. We map it to |${(V,H,M,{\textit{KB}},k+1)}$|⁠, where $$\begin{align*} {\textit{KB}} &:= \{(\neg x \lor \neg y) \mid (x,y) \in \tilde E\},\\ H &:= \textrm{var}({\textit{KB}}) \cup \{z\},\\ M &:= z. \end{align*}$$ Let |$U$| be an independent set of size |$k$| then |$U\land{\textit{KB}}$| is consistent because no two elements with an edge are in |$U$|⁠. As a consequence, |$U\cup \{z\}$| is an explanation for |${(V,H,M,{\textit{KB}},k+1)}$|⁠. Conversely, an explanation |$E$| for |${(V,H,M,{\textit{KB}},k+1)}$| of size |$k+1$| must include |$z$| as well as |$k$| other variables. Now, |$E \land{\textit{KB}}$| is consistent and this implies that no variables in |$E$| have an edge, consequently giving an independent set of size |$k$|⁠. This implies that |$(V, \tilde E)$| admits an independent set of size |$k$| if and only if |${(V,H,M,{\textit{KB}})}$| admits an explanation of size |$k+1$|⁠. Now we prove |$\textbf{W}[1]$| membership for |$\textrm{ABD}_=(S, |E|)$| with |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{12}}$|⁠, this means also arbitrary bases, in the lemma below. Lemma 3.23 Let |$\ell \ge 2$|⁠, then |$\textrm{ABD}_=(S, |E|)$| is in |$\textbf{W}[1]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{12}}$|⁠. Proof. We reduce |$\textrm{ABD}_=(S, |E|)$| to |${\textrm{p-WSAT}(\varGamma _{1,\ell })}$|⁠, which is |$\textbf{W}[1]$|-complete due to Prop. 2.4. Note that |$\varGamma _{1,\ell }$| is the class of |$\ell $|-CNF formulas. We want to mention here that the proof is correct even in the presence of equality constraints. As a consequence, the base independence is not implied by any of the previous lemmas but it follows due the proof below. According to Lemma 3.13, we can determine whether there exists a solution of size |$\leq s$| in polynomial time. Let |${(V,H,M,{\textit{KB}},k)}$| be an instance of |$\textrm{ABD}_=({\textsf{IS}^\ell_{12}}, |E|)$| with |${\textit{KB}} = \bigwedge _{i\leq r} C_i \land N \land P \land E$|⁠, where |$C_i = (\neg{x^i_1} \lor \dots \lor \neg{x^i_\ell })$|⁠, and |$P, N$| denote the positive and negative unit clauses, respectively, and |$E$| are the equality clauses. Without loss of generality, assume that |${(V,H,M,{\textit{KB}},k)}$| admits a solution of size |$\leq k$| (otherwise, map it to a negative dummy instance). Moreover, it follows from Lemma 3.13 that in this case any solution |$E$| satisfies that |$E_{MP} \subseteq E \subseteq H$|⁠. This implies |$s \geq |E_{MP}|$|⁠. We also know from Lemma 3.13 that |$E_{MP}$| is an explanation for |$M$| and that both |$E_{MP}$| and |$M$| are consistent with all clauses in |${\textit{KB}}$|⁠. The question now reduces to whether we can extend |$E_{MP}$| to a solution of size |$k$| by adding |$k - |E_{MP}|$| variables from |$H \setminus E_{MP}$|? We show that this can be achieved and map |${(V,H,M,{\textit{KB}},k)}$| to |$\langle \varphi , k - |E_{MP}|\rangle $|⁠, where |$\varphi $| is obtained from |${\textit{KB}}$| by the following consecutive steps: first we take care of equality clauses. For each |$x_i=x_j \in{\textit{KB}}$|⁠, such that |$x_i, x_j \in H$|⁠, add to |$\phi $| the clauses |$(\neg x_i\lor x_j)$| and |$( x_i\lor \neg x_j)$|⁠. We add these two clauses to |$\phi $| ensuring that corresponding to each clause of the form |$x_i=x_j$|⁠, either both |$x_i, x_j$| are in the solution, or none is. 1. Remove all clauses |$C_i$| containing only variables not from |$H$|⁠. Remove all negative unit clauses |$(\neg x)\in N$| such that |$x \notin H$|⁠. Note that after this step all remaining negative unit clauses are built upon variables from |$H \setminus E_{MP}$| only. For each clause |$C_i$|⁠, denote by |$X^i_H$| (resp., |$X^i_{\overline{H}}$|⁠) the variables from |$H$| (resp., not from |$H$|⁠). Execute the following: (a) Remove |$C_i$|⁠. (b) If |$X^i_{\overline{H}} \subseteq P$|⁠: add to |$\varphi $| the clause |$(\neg x_1 \lor \dots \lor \neg x_p)$|⁠, where |$\{x_1, \dots , x_p\} = X^i_H \setminus E_{MP}$|⁠. Otherwise nothing needs to be done as |$X^i_{\overline{H}}\not \subseteq P$| is true. Then, for some variable |$x\notin P$| we have that |$\lnot x\lor \bigvee _{x_j\in X^i_H}\lnot x_j$| is satisfiable via setting |$x$| to |$0$| if all |$x_j$| are mapped to |$1$|⁠. Note that after this step all remaining clauses |$C_i$| are built upon variables from |$H$| only. Remove all positive unit clauses |$(x) \in P$| such that |$x \notin H \setminus E_{MP}$|⁠. Note that after this step it holds that |$\textrm{var}(\varphi ) = H$| and all remaining positive unit clauses are built upon variables from |$H \setminus E_{MP}$| only. For all clauses |$C_i$|⁠: remove from |$C_i$| all literals built upon variables from |$E_{MP}$|⁠. Note that in the so obtained |$C_i^{\prime}$| at least one literal remains, because otherwise |$E_{MP}$| would be inconsistent with |$C_i$|⁠. After the last step has been implemented, it holds that |$\textrm{var}(\varphi ) = H \setminus E_{MP}$|⁠. As a consequence, the following equivalences are true: – |${(V,H,M,{\textit{KB}},k)} \textrm{ admits a solution of size exactly} k$|⁠. |$E_{MP}$| extends to a solution of size |$k$| by adding |$k - |E_{MP}|$| variables. |$\varphi $| has a satisfying assignment of size exactly |$k - |E_{MP}|$|⁠. This completes the proof to the lemma. 3.5 Parameter ‘number of manifestations’ |$|M|$| The complexity landscape regarding the parameter |$|M|$| is more diverse. The classification differs for each of the investigated problem variants. Consequently, we treat each case separately and start with the general abduction problem that provides a pentachotomy. Theoem 3.24 The problem |$\textrm{ABD}(S, |M|)$| is 1. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, 2. |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, 3. |${\textbf{para-}}{\textbf{NP}} $|-complete if |${\textsf{IE}}_0\subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IE}}_2 $|⁠, 4. |$\textbf{W}[1]$|-complete if |${\textsf{IS}^{2}_{11}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{C}}$| and |${\textsf{C}} \in \{{\textsf{ID}}_2, {\textsf{IS}^{\ell }_{10}}\}$|⁠, 5. |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in \{{\textsf{ID}}_1, {\textsf{IS}_{12}},{\textsf{IE}}_1,{\textsf{IV}}_2\}$|⁠. Proof. (1.+2.) We proved in Lemma 3.9 using the fact that 1-slice of each problem is hard for respective classes. (3.) The membership is trivial since the classical problem is NP-complete. For hardness, we prove that |$1$|-slice of the problem is NP-hard. Notice that due to Nordh and Zanuttini [31, Lemma 29] an abduction instance can be reduced to an instance with only one manifestation if the KB allows certain clauses. The idea is to encode a set of manifestations |$M$| by a single new manifestation |$y$|⁠, while adding the clause |$y\lor \bigvee \limits _{m\in M} \neg m$| to the |${\textit{KB}}$|⁠. Recall that |$M$| is a (positive) set of propositions, implying that the clause |$y\lor \bigvee \limits _{m\in M} \neg m$| is a Horn clause. Consequently, the aforementioned reduction is valid for |${\textit{KB}} \in{\textsf{IE}}_0$|⁠. This reduction in conjunction with the result for single literal manifestation ([31, Prop. 53]) implies that |$1$|-slice of the problem |$\textrm{ABD}(\textsf{IE}_0, |M|)$| is |${\textbf{NP}} $|-hard. Consequently, the problem is |${\textbf{para-}}{\textbf{NP}} $|-complete. (4.) The membership for |${S}$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_2$| follows from [21, Thm. 25]. Notice that the authors in [21] prove the completeness for the languages in |${\textsf{ID}}_2$| alone, but using the fact that the formulas (or clauses) in their reduction are |${\textsf{IS}^{2}_{11}}$|-formulas, we derive the hardness for |${\textsf{IS}^{2}_{11}}$|⁠. The |$\textbf{W}[1]$|-membership for the languages |${S}$| such that |$ \left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$| follows from Lemma 3.27. As a consequence, we have the desired completeness results. (5.) Follows from Corollary 3.10. For |$\textrm{ABD}_\leq$|⁠, definite Horn cases surprisingly behave different and are much harder than for the general case. Theoem 3.25 The problem |$\textrm{ABD}_leq(S, |M|)$| is 1. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, 2. |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, 3. |${\textbf{para-}}{\textbf{NP}} $|-complete if |${\textsf{IE}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{IE}}_2$|⁠, 4. |$\textbf{W}[1]$|-complete if |${\textsf{IS}^{2}_{11}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{C}}$| and |${\textsf{C}} \in \{{\textsf{ID}}_2, {\textsf{IS}^{\ell }_{10}}\}$|⁠, 5. |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}\in \{{\textsf{ID}}_1, {\textsf{IS}_{12}}, {\textsf{IV}}_2\}$|⁠. Proof. (1.+2.) Follows from Theorem 3.24 in conjunction with Lemma 3.7. (3.) The membership is trivial since classical problem is in NP. For hardness, we reduce VertexCover to our problem similar to the approach of Fellows et al. [21, Thm. 5]. The problem can be translated into an abduction instance with |${\textit{KB}}\in{\textsf{IE}}$|⁠, consequently giving the desired hardness result. (4.) The membership for |${S}$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{ID}}_2$| follows from [21, Thm. 25]. Notice that the authors in [21] prove the completeness for the languages in |${\textsf{ID}}_2$| alone, but using the fact that the formulas (or clauses) in their reduction are |${\textsf{IS}^{2}_{11}}$|-formulas, we derive the hardness for |${\textsf{IS}^{2}_{11}}$|⁠. The |$\textbf{W}[1]$|-membership for the languages |${S}$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$| follows from Lemma 3.28. (5.)FPT-membership for |$\textrm{ABD}_leq(S, |M|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IM}}$| follows from Lemma 3.29. Lemma 3.30 extends this result to |$\textrm{ABD}_leq(S, |M|)$| such that |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. The remaining cases are due to Corollary 3.15. Now, we end by stating the results for |$\textrm{ABD}_=$|⁠. Interesting to observe, the majority of the intractable cases is already much harder with large parts being |${\textbf{para-}}{\textbf{NP}} $|-complete. Even the case of the essentially negative co-clones that are FPT for |$\textrm{ABD}_\leq $| yield |${\textbf{para-}}{\textbf{NP}} $|-completeness in this situation. Merely the |$2$|-affine and dualHorn cases are |${\textbf{FPT}} $|⁠. Theoem 3.26 The problem |$\textrm{ABD}_=(S, |M|)$| is 1. |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-hard if |${\textsf{C}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{BR}}$| and |${\textsf{C}}\in \{{\textsf{IN}}_2,{\textsf{II}}_0\}$|⁠, 2. |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard if |${\textsf{IN}} \subseteq \left \langle{S} \right \rangle \subseteq{\textsf{II}}_1$|⁠, 3. |${\textbf{para-}}{\textbf{NP}} $|-complete if |${\textsf{C}}_1 \subseteq \left \langle{S} \right \rangle $| where |${\textsf{C}}_1\in \{{\textsf{IS}^{2}_{1}},{\textsf{IE}}\}$| and |$\left \langle{S} \right \rangle \subseteq{\textsf{C}}_2 \in \{{\textsf{IE}}_2, {\textsf{ID}}_2\}$|⁠, 4. |${\textbf{FPT}} $| if |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in \{{\textsf{ID}}_1, {\textsf{IV}}_2\}$|⁠. Proof. (1.+2.) Follow from Theorem 3.24 in conjunction with Lemma 3.7. (3.) The membership in each case is trivial since the classical problems are NP-complete. The hardness for |${\textsf{IE}}\subseteq \left \langle{S} \right \rangle $| follows from the argument used in the proof of Theorem 3.25 for the |${\textsf{IE}}$| case. The hardness for the remaining cases follows from Lemma 3.31 where we prove that the problem |$\textrm{ABD}_=(S, |M|)$| is |${\textbf{para-}}{\textbf{NP}} $|-hard as long as |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠. The case for |$\textrm{ABD}_=({\textsf{IS}^{2}_{1}}, |M|)$|⁠, so also arbitrary bases, then follows as a corollary. (4.) The proof for |${\textsf{IV}}_2$| is due to the monotone argument of Lemma 3.8 and Theorem 3.25. The result for |${\textsf{ID}}_1$| is due to Corollary 3.15. 3.5.1 Intermediate lemmas Lemma 3.27 Let |$\ell \geq 2$| then the problem |$\textrm{ABD}(S, |M|)$| is in |$\textbf{W}[1]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. We use the same reduction as in Lemma 3.21 and argue that the so-obtained multi-tape Turing machine |$\mathbb M$| can be simulated by a single-tape machine |$\mathbb M^{\prime}$|⁠. This gives a reduction to the |$\textbf{W}[1]$|-complete problem |${\textrm{Short-NSTM-Halt}}$| [22, Thm. 6.17]. Moreover, the blow-up in the size of |$\mathbb M^{\prime}$| is only a function in the parameter |$k$| and |$\mathbb M^{\prime}$| runs for a number of steps that is bounded by a function in |$k$|⁠. In |$\textrm{ABD}$|⁠, there is no restriction on the solution size. As a consequence, the abduction instance has a solution if and only if the formula |$\psi $| of the reduction in Lemma 3.21 is satisfiable. Recall that the number of positive clauses in |$\psi $| is |$|M|$|⁠, which is the parameter |$k$| in this case. We claim that it is enough to determine if |$\psi $| has a satisfying assignment of weight at most |$k$|⁠. Claim. Let |$\phi $| be a |$\varGamma _{2,1}$|-formula with |$k$| positive and |$n$| negative clauses. Then |$\phi $| is satisfiable if and only if |$\phi $| has a satisfying assignment of size at most |$k$|⁠. Proof of Claim. The direction from right to left is trivial. For the other direction, note that if |$s\models \phi $| for some assignment |$s$| then, |$s \cap U_i \not = \emptyset $| for any |$i\leq k$| and |$ V_j \not \subseteq s $| for |$j\leq N$|⁠. Where we consider |$s$| as the collection of variables mapped to |$1$|⁠. Let |$s^{\prime}$| be the assignment obtained from |$s$| such that, for each positive clause |$U_i$|⁠, |$s^{\prime}$| selects exactly one variable from |$U_i$| (with repetition allowed for different clauses). Then |$s^{\prime}\models \phi $| and |$|s^{\prime}|\leq k$|⁠. This is because |$s^{\prime}$| selects exactly one variable from each positive clause and |$s^{\prime}\subseteq s$|⁠. As before, we ignore the negative clauses of length more than |$k$|⁠. This implies that there can be at most |$2^k$| negative clauses. Consequently, |$\mathbb M$| has at most |$k+2^k+1$| tapes. We argue that the size of |$\mathbb M^{\prime}$| is |$O(2^k\cdot k\cdot p(|\psi |))$| where |$p$| is some polynomial. This is because there are |$2^k+k+1$| tapes (the worst case) and consequently, the size of each transition is bounded by |$O(k+2^k)$|⁠. This implies that the size of |$\mathbb M^{\prime}$| is |$O((k+2^k)\cdot |\mathbb M|)$| where |$|\mathbb M|= O(k \cdot |\psi |^2)$|⁠. Moreover, |$\mathbb M^{\prime}$| runs for |$2^k\cdot f(k)^2$| steps where |$f(k)$| is the number of steps taken by |$\mathbb M$| (for details of the simulation see the textbook of Sipser [43, Theorem 7.8]). The correctness follows from Lemma 3.21. This completes the proof of the lemma. Lemma 3.28 Let |$\ell \geq 2$| then the problem |$\textrm{ABD}_\leq(S, |M|)$| is in |$\textbf{W}[1]$| if |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Proof. Given |${(V,H,M,{\textit{KB}},s,|M|)}$|⁠, the task is to find an explanation of size at most |$s$| where |$|M|=k$| is the parameter. We argue that the reduction in Lemma 3.27 can be extended to this case. In Lemma 3.27 we proved that there is an explanation for |$\textrm{ABD}(S, |M|)$| if and only if there is an explanation of size |$|M|$| at most, where |$\left \langle{S} \right \rangle \subseteq{\textsf{IS}^{\ell }_{10}}$|⁠. Now we have two cases. If |$k \leq s$| then the result holds already due to Lemma 3.27 (in particular the claim). This is because a solution of size |$k$| is also a solution of size at most |$s$| and there can be no solution of size in between |$k$| and |$s$| if there is no solution of size at most |$k$|⁠. On the other hand, if |$s< k$|⁠, the solution size is still bounded by the parameter. Our reduction in the proof of Lemma 3.27 takes care of this change by producing two parameters |$k_1$| and |$k_2$| such that |$k_1=k=|M|$| and |$k_2=s$|⁠. This refined reduction is still an FPT-reduction because the parameter |$f(k)=k_1+k_2$| is bounded by |$k$|⁠, i.e. |$f(k)\leq 2k$|⁠. The rest of the reduction remains the same. The only difference now is that the machine (say |$\mathbb M^{\prime\prime}$|⁠) guesses |$k_2$| elements and not |$k_1$| as |$\mathbb M^{\prime}$| does in Lemma 3.27. This completes the proof to the current lemma. We prove FPT-membership for |$\textrm{ABD}_\leq(S, |M|)$|⁠, where |$\left \langle{S} \right \rangle \subseteq{\textsf{IM}}$|⁠, by reducing our problem to the |${\textrm{MaxSATs}}$| problem that asks, given |$m$| clauses, is it possible to set at most |$s$| variables to true so that at least |$k$| clauses are satisfied. Lemma 3.29 The problem |$\textrm{ABD}_\leq(S, |M|)$| is FPT if |$\left \langle{S} \right \rangle \subseteq{\textsf{IM}}$|⁠. Proof. Given an instance |${(V,H,M,{\textit{KB}},s,|M|)}$| with |${\textit{KB}}=\bigwedge \limits _{i\leq r}(x_i \rightarrow y_i)$| and |$M= m_1\land \ldots \land m_{|M|}$|⁠. Recall that each |$m_i\in M$| can be explained by a single |$h_i\in H $|⁠. If |$|M|\leq s$| then there is nothing to prove. This is due the fact that in the proof of Lemma 3.18, there are fewer than |$s$| many sets of the form |$H_i$| each explaining an |$m_i\in M$|⁠. As a consequence, we need only select one |$h_{i,j}$| from each |$H_i$| as the part of a solution to yield a solution of size |$\leq s$|⁠. Accordingly, assume that |$|M|> s$|⁠. Proceed as in the proof of Lemma 3.18 and associate a set |$H_i \subseteq H$| of hypotheses with each |$m_i$| that explains it for |$i\leq |M|$|⁠. It is enough to check whether selecting at most |$s$| many elements |$h_i\in H$| can explain all the manifestations |$m_i \in M$|⁠. We reduce our problem to |${\textrm{MaxSATs}}$| [5] (we alter the notation slightly) asking, given a |$\textrm{CNF}$| formula on |$n$| variables with |$m$| clauses, if setting at most |$s$| variables to true satisfies at least |$k$| clauses. Let |$H^{\prime}$| be the collection of all |$H_i$|’s. For each |$i$| let |$C_i$| be the clause |$\bigvee \limits _{j} h_j$| where |$h_j\in H_i$|⁠. Furthermore, let |$C$| be the collection of all such clauses. Then |$C$| is built over variables in |$V^{\prime} = \bigcup _i H_i$|⁠. Our reduction maps |${(V,H,M,{\textit{KB}},s,|M|)}$| to |$\langle C, s, k \rangle $|⁠. Note that we only have |$|M|=k$| many clauses in |$C$| and, as a result, the question reduces to whether it is possible to set at most |$s$| variables from |$V^{\prime}$| to satisfy every clause in |$C$|? The reduced problem |${\textrm{MaxSATs}}$| when parameterized by |$k$| (the minimum number of clauses to be satisfied) is |${\textbf{FPT}} $| [5, Prop. 4.3]. Now, each |$H_i$| can be computed in polynomial time, the whole computation is a polynomial time reduction. Finally, the new parameter value |$k$| is exactly the same as the old parameter |$|M|$|⁠, the reduction is an |$\leq ^\textbf{FPT}$|-reduction. As a consequence, the lemma applies. We extend the |${\textbf{FPT}} $|-membership to languages in |${\textsf{IV}}_2$| using the same argument as in Lemma 3.19. Corollary 3.30 The problem |$\textrm{ABD}_\leq(S, |M|)$| is FPT if |$\left \langle{S} \right \rangle \subseteq{\textsf{IV}}_2$|⁠. Proof. After applying unit propagation and resolution we can ignore the positive clauses of length |$\geq 2$| and clauses with one negative literal of length |$\geq 3$|⁠. Lemma 3.31 For any constraint language |${S}$| such that |$\neg x\lor \neg y \in \left \langle{S} \right \rangle _{\neq }$|⁠, the problem |$\textrm{ABD}_=(S, |E|)$| is |${\textbf{para-}}{\textbf{NP}} $|-hard. Proof. We prove that the |$1$|-slice of the problem is |${\textbf{NP}} $|-hard by reducing from classical |${\textrm{IndependentSet}} $| (which is |${\textbf{NP}} $|-complete [25]) to |$\textrm{ABD}_=(S)$|⁠. The reduction is essentially the classical counterpart of the one presented in Lemma 3.22. Let |$\langle V, \tilde E\rangle $| be an instance of |${\textrm{IndependentSet}} $|⁠. We map it to |${(V,H,M,{\textit{KB}},s)}$|⁠, where $$\begin{align*} {\textit{KB}} &:= \{(\neg x \lor \neg y) \mid (x,y) \in \tilde E\},\\ H &:= \textrm{var}({\textit{KB}}) \cup \{z\},\\ M &:= z,\\ s &:= k+1. \end{align*}$$ Then |$(V, \tilde E)$| admits an independent set of size |$k$| if and only if |${(V,H,M,{\textit{KB}},s)}$| admits an explanation of size |$s$|⁠. Notice that we did not mention the base independence for essentially negative languages in the previous proof. This is because, the |${\textbf{para-}}{\textbf{NP}} $|-membership as well as the base independence holds for |$\textrm{ABD}_=(S, |M|)$|⁠, where |$\left \langle{S} \right \rangle \subseteq{\textsf{C}} \in \{{\textsf{IE}}_2,{\textsf{ID}}_2\}$|⁠. This gives the desired results for essentially negative languages as well. 4 Conclusion In this paper, we presented a two-dimensional classification of three central abductive reasoning problems (unrestricted explanation size, = and |$\leq $|⁠). In one dimension, we consider the different parameterizations |$|H|,|M|,|V|,|E|$|⁠, and in the other dimension we consider all possible constraint languages defined by corresponding co-clones except the affine co-clones. Often in the past, problems regarding the affine co-clones (resp., clones) resisted a complete classification [1, 2, 11, 13, 23, 27, 38, 44]. Also the result of Durand and Hermann [19] underlines how restive problems around affine functions are. It is difficult to explain why exactly these cases are so problematic but the notion of the Fourier expansion [32] of Boolean functions gives a nice and fitting view on that. Informally, the Fourier expansion of a Boolean function is a probability measure mimicking how likely a flip of a variable changes the function value. For instance, disjunctions have a very low Fourier expansion value, whereas the exclusive-or function has the maximum. Affine functions can though be seen as rather counterintuitive as every variable influences the function value dramatically. For all three studied problems, we exhibit the same trichotomy for the parameter |$|H|$| (⁠|${\textsf{IN}}$| is |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard, |${\textsf{IN}}_2$| is |${\textbf{para-}}{\textbf{DP}} $|-hard and the remaining are |${\textbf{FPT}} $|⁠). The parameter |$|V|$| always allows for FPT algorithms independent of the co-clone. Regarding |$|E|$|⁠, only the two size restricted variants are meaningful. For ‘|$\leq $|’ we achieve a pentachotomy between |${\textbf{FPT}} $|⁠, |$\textbf{W}[2]$|-complete, |$\textbf{W}[\textbf{P}] $|-complete, |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|- and |${\textbf{para-}}{\textbf{DP}} $|-hard. Whereas, for ‘=’, we achieve a hexachotomy additionally having |$\textbf{W}[1]$|-completeness for the essentially negative cases. These |$\textbf{W}[1]$|-hard cases are also surprising in the sense that for ‘|$\leq $|’ they are easy and |${\textbf{FPT}} $|⁠. Similarly, the same easy/hard-difference has been observed as well for |$|M|$| as the studied parameter. However, here, we distinguish between |${\textbf{para-}}{\textbf{NP}} $|-complete for ‘=’ and |${\textbf{FPT}} $| for ‘|$\leq $|’. The complete picture for ‘=’ and |$|M|$| is a tetrachotomy ranging through |${\textbf{FPT}} $|⁠, |${\textbf{para-}}{\textbf{NP}} $|-complete, |${\textbf{para-}}{\textbf{co}}{\textbf{NP}} $|-hard and |${\textbf{para-}}{\boldsymbol{\varSigma _2^P}} $|-complete. Also, parameterized enumeration complexity [7, 8] is the next object of our investigations. Here, we already made some initial observations yielding |${\textbf{FPT-enum}}$|-algorithms for |$|V|$| and |${\textsf{BR}}$| as well as for |$|H|$| and |${\textsf{IE}}_2,{\textsf{IV}}_2,{\textsf{ID}}_2$| and |${\textsf{IL}}_2$|⁠. Such algorithms produce the whole set of solutions in |${\textbf{FPT}} $|-time. Furthermore, |${\textsf{IL}}_1$| even allows for |${\textbf{FPT}}$|-algorithms for any parameterization for the problem |$\textrm{ABD}(S, ks)$| (so it extends Corollary 3.10 in that way). Notice that in this paper, we did not require |$H\cap M$| to be empty as, for instance, Fellows et al. [21] assumed. 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