1. Introduction
By Schaefer’s famous theorem [24], we know that the SAT problem becomes tractable under certain syntactic restrictions such as the restriction to Horn formulas (i.e., formulas in CNF where clauses have at most one positive literal) or to Krom formulas (i.e., clauses have at most two literals). Propositional formulas play an important role in reasoning problems in a great variety of areas such as belief change, logicbased abduction, closedworld reasoning, etc. Most of the relevant problems in these areas are intractable. It is therefore a natural question whether restrictions on the formulas involved help to decrease the high complexity. While the restriction to Horn formulas is usually well studied, other restrictions in Schaefer’s framework have received much less attention. In this work, we have a closer look at the restriction to Krom formulas and its effect on the complexity of several hard reasoning problems from the AI domain.
One particular source of complexity of such problems is the involvement of some notion of minimality. In closed world reasoning, we add negative literals or more general formulas to a theory only if they are true in every minimal model of the theory in order to circumvent inconsistency. In belief revision we want to revise a given belief set by some new belief. To this end, we retain only those models of the new belief which have minimal distance to the models of the given belief set. In abduction, we search for a subset of the hypotheses (i.e., an “explanation”) that is consistent with the given theory and which – together with the theory – explains (i.e., logically entails) all manifestations. Again, one is usually not content with any subset of the hypotheses but with a minimal one. However, different notions of minimality might be considered, in particular, minimality w.r.t. set inclusion or w.r.t. cardinality. In abduction, these two notions are directly applied to explanations [9]. In belief revision, one is interested in the models of the new belief which have minimal distance from the models of the given belief set. Distance between models is defined here via the symmetric set difference of the atoms assigned to true in the compared models. Dalal’s revision operator [7] seeks to minimize the cardinality of while Satoh’s operator [23] defines the minimality of in terms of set inclusion.
The chosen notion of minimality usually has a significant impact on the complexity of the resulting reasoning tasks. If minimality is defined in terms of cardinality, we often get problems that are complete for some class with (which is also referred to as , see [27]). For instance, two of the most common reasoning tasks in belief revision (i.e., model checking and implication) are complete for Dalal’s revision operator [8, 17]. Abduction aiming at cardinalityminimal explanations is complete [9]. If minimality is defined in terms of set inclusion, completeness results for one of the classes or for some are more common. For instance, belief revision with Satoh’s revision operator becomes complete (for model checking) [17] respectively complete (for implication) [8]. Similarly, abduction is complete if we test whether some hypothesis is contained in a subsetminimal explanation [9].
For the above mentioned problems, various ways to decrease the complexity have been studied. Indeed, in almost all of these cases, a restriction of the involved formulas to Horn makes the complexity drop by one level in the polynomial hierarchy. Only belief revision with Dalal’s revision operator remains complete in the Horn case, see [8, 17, 9]. The restriction to Krom has not been considered yet for these problems. In this paper we show that the picture is very similar to the Horn case. Indeed, for the considered problems in belief revision and abduction, we get for the Krom case exactly the same complexity classifications. Actually, we choose the problem reductions for our hardness proofs in such a way that we thus also strengthen the previous hardness results by showing that they even hold if formulas are restricted to Horn and Krom at the same time.
To get a deeper understanding why certain problems remain hard in the Krom case, we have a closer look at a related variant of the SAT problem, where also cardinality minimality is involved. We define the cardinality of a truth assignment (or a model) as the cardinality of the set of variables assigned true. Thus, we consider the CardMinSat problem: given a propositional formula and an atom , is true in some cardinalityminimal model of ? It is easy to show that this problem is complete. If is restricted to Horn, then this problem becomes trivial, since Horn formulas have a unique minimal model which can be efficiently computed. But what happens if we restrict to Krom? We show that completeness holds also in this case. This hardness result will then also be very convenient for proving our hardness results for the considered problems in belief revision and abduction. Since CardMinSat seems to be central in evaluating the complexity of reasoning problems in which some kind of cardinalityminimality is involved we investigate its complexity in a deeper way, in particular by characterizing the tractable cases.
The main contributions of the paper are as follows:

Prototypical problems for and . In Section 3, we first review the prototypical problem LexMaxSat. Hardness for this problem is due to [15] but only follows implicitly from that work. We reformulate the proofs in terms of standard terminology in order to explicitly prove completeness for the related problem LogLexMaxSat in an analogous way. Note that LogLexMaxSat, which will be the basis for our forthcoming results, is not mentioned explicitly in [15].

SAT variants. In Section 4, we investigate the complexity of the CardMinSat problem and the analogously defined CardMaxSat problem. Our central result is that these problems remain complete for formulas in Krom form.

Applications. We investigate several reasoning problems in the areas of belief revision and abduction in Section 5. For the restriction to Krom form, we establish the same complexity classifications as for the previously known restriction to Horn form. In fact, we thus also strengthen the previously known hardness results by showing that hardness even holds for the simultaneous restriction to Horn and Krom.

Classification of CardMinSat inside the Krom fragment. In Section 6 we investigate the complexity of CardMinSat within the Krom fragment in order to identify the necessary additional syntactic restrictions towards tractability. To this aim we use the wellknown framework by Schaefer and obtain a complete complexity classification of the problem.
2. Preliminaries
Propositional logic. We assume familiarity with the basics of propositional logic [1]. We only want to fix some notation and conventions here. We say that a formula is in CNF (resp. DNF) for if all its clauses (resp. terms) have at most literals. Formulas in CNF are also called Krom. A formula is called Horn (resp. dual Horn
) if it is in CNF and in each clause at most one literal is positive (resp. negative). It is convenient to identify truth assignments with the set of variables which are true in an assignment. It is thus possible to consider the cardinality and the subsetrelation on the models of a formula. If in addition an order is defined on the variables, we may alternatively identify truth assignments with bit vectors, where we encode true (resp. false) by 1 (resp. 0) and arrange the propositional variables in decreasing order. We thus naturally get a lexicographical order on truth assignments.
Complexity Classes. All complexity results in this paper refer to classes in the Polynomial Hierarchy (PH) [21]. The building blocks of PH are the classes and of decision problems solvable in deterministic resp. nondeterministic polynomial time. The classes , , and of PH are inductively defined as and , , , where we write (resp.
) for the class of decision problems that can be decided by a deterministic (resp. nondeterministic) Turing machine in polynomial time using an oracle for the class
. The prototypical complete problem is QSAT, (i.e., quantified satisfiability with alternating blocks of quantifiers, starting with ), where we have to decide the satisfiability of a formula (withfor odd
and for even ) with no free propositional variables. W.l.o.g., one may assume here that is in 3DNF (if ), resp. in 3CNF (if ). In [27], several restrictions on the oracle calls in a computation have been studied. If on input with at most calls to the oracles are allowed, then we get the class which is also referred to as . A large collection of complete problems is given in [15, 11, 10]. Sections 3 and 4 in this work will also be devoted to completeness results. Several problems complete for with can be found in [16]. Hardness results in all these classes are obtained via logspace reductions, and we write when a problem is logspace reducible to a problem .3.  and completeness
In this work, we prove several new  and completeness results. Problems in these classes naturally arise from optimization variants of complete problems, where a sequence of oracle calls for the underlying complete problem is required to compute the optimum. completeness applies, if binary search with logarithmically many oracle calls suffices to find the optimum. Recently, an additional intuition of complete (and, more generally, complete) problems has been presented in [18], namely counting the number of positive instances in a set of instances of some complete (or, more generally, complete) problem. However, the problems for which we establish  and completeness here fall into the category of optimization problems derived from complete problems.
The base problems for proving our new  and completeness results are the following: The following problems are considered as prototypical for the classes and . In particular, hardness of LogLexMaxSat will be used as starting point for our reductions towards hardness of CardMinSat for Krom formulas in Section 4.

LexMaxSat

Propositional formula and an order on the variables in .

Is true in the lexicographically maximal model of ?

LogLexMaxSat

Propositional formula and an order on some of the variables in with .

Is true in the lexicographically maximal bit vector that can be extended to a model of ?
The LexMaxSat problem will serve as our prototypical complete problem while the LogLexMaxSat problem will be the basis of our completeness proofs. The completeness of LexMaxSat is stated in [15] – without proof though (see Theorem 3.4 in [15]). The membership is easy. For the hardness, the proof is implicit in a sequence of lemmas and theorems in [15]. However, the main goal in [15] is to advocate a new machine model (socalled metric Turing machines) for defining new complexity classes of optimization problems (the socalled and classes). The LogLexMaxSat problem is not mentioned explicitly in [15] but, of course, it is analogous to the LexMaxSat problem.
To free the reader from the burden of tracing the line of argumentation in [15] via several lemmas and theorems on the and classes, we give direct proofs of the  and completeness of LexMaxSat and LogLexMaxSat, respectively, in the standard terminology of oracle Turing machines (cf. [21]). In the first place, we thus have to establish the connection between oracle calls and optimization. To this end, we introduce the following problems:

NPMax

Turing machine running in nondeterministic polynomial time and producing a binary string as output; string as input to .

Let denote the lexicographically maximal output string over all computation paths of on input ; does the last bit of have value 1?

LogNPMax

Turing machine running in nondeterministic polynomial time and producing a binary string, whose length is logarithmically bounded in the size of the Turing machine and the input; string as input to .

Let denote the lexicographically maximal output string over all computation paths of on input ; does the last bit of have value 1?
The NPMax problem is complete and the LogNPMax problem is complete.
Proof.
The membership of NPMax and the membership of LogNPMax is seen by the following algorithm, which runs in deterministic polynomial time and has access to an oracle. The algorithm maintains a bit vector of the lexicographically maximal prefix of possible outputs of TM on input . To this end, we initialize to and ask the following kind of questions to an oracle: Does there exist a computation path of TM on input , such that the first output bits are and outputs yet another bit? If the answer to this oracle call is “no” then the algorithm stops with acceptance if ( and ) holds and it stops with rejection if ( or ) holds.
If the oracle call yields a “yes” answer, then our algorithm calls another oracle with the question: Does there exist a computation path of TM on input , such that the first output bits are ? If the answer to this oracle call is “yes”, then we set ; otherwise we set . In either case, we then increment by 1 and continue with the first oracle question (i.e., does there exist a computation path of TM on input , such that the first output bits are and outputs yet another bit?).
Suppose that the lexicographically maximal output produced by on input has bits. Then our algorithm needs in total oracle calls and the oracles work in nondeterministic polynomial time. Moreover, if the size of the output string of is logarithmically bounded, then the number of oracle calls is logarithmically bounded as well.
For the hardness part, we first concentrate on the NPMax problem and discuss afterwards the modifications required to prove the corresponding complexity result also for the LogNPMax problem. Consider an arbitrary problem in , i.e., is decided in deterministic polynomial time by a Turing machine with access to an oracle for the SATproblem.
Now let be an arbitrary instance of problem . From this we construct an instance of NPMax, where we leave unchanged and we define as follows: In principle, simulates the execution of on input . However, whenever reaches a call to the SAToracle, with some input say, then nondeterministically executes on . In other words, in the computation tree of , the subtree corresponding to this nondeterministic execution of on is precisely the computation tree of on . On every computation path ending in acceptance (resp. rejection) of , the TM writes 1 (resp. 0) to the output. After that, continues with the execution of as if it had received a “yes” (resp. a “no”) answer from the oracle call. After the last oracle call, executes to the end. If ends in acceptance, then outputs 1; otherwise it outputs 0 as the final bit.
It remains to prove the following claims:

Correctness. Let denote the lexicographically maximal output string over all computation paths of on input . Then the last bit of has value 1 if and only if is a positive instance of .

Polynomial time. The total length of each computation path of on input is polynomially bounded in .

Logarithmically bounded output. If the number of oracle calls of TM is logarithmically bounded in its input (i.e., problem is in ), then the size of the output produced by on input is also logarithmically bounded.
Correctness. We first have to prove the following claim: for every and for every bit vector of length : is a prefix of the lexicographically maximal output string over all computation paths of on input if and only if encodes the correct answers of the first oracle calls of TM on input , i.e., for , (resp. ) if the th oracle call yields a “yes” (resp. a “no”) answer. This claim can be easily verified by induction on . Consider the induction begin with . We have to show that the first bit of the output string is 1 if and only if the first oracle call yields a “yes” answer. Indeed, suppose that the first oracle does yield a “yes” answer. This means that at least one of the computation paths of the nondeterministic oracle machine ends with acceptance. By our construction of , there exists at least one computation path of the nondeterministic TM on which value 1 is written as first bit to the output. Hence, the string ’1’ is indeed the prefix of length 1 of the lexicographically maximal output string over all computation paths of on input . Conversely, suppose that the first oracle call yields a “no” answer. This means that all of the computation paths of the nondeterministic oracle machine end with rejection. By our construction of , all computation paths of the nondeterministic TM write 0 as the first bit to the output. Hence, the string ’0’ is indeed the prefix of length 1 of the lexicographically maximal output string. The induction step is shown by the same kind of reasoning.
Now let denote the number of oracle calls carried out by TM on input and let denote the lexicographically maximal output of TM over all its computation paths. Then has length such that the first bits encode the correct answers of the oracle calls of TM on input . Moreover, by the construction of , we indeed have that the last bit of is 1 (resp. 0), if and only if is a positive (resp. negative) instance of .
Polynomial time. Suppose that on input with is guaranteed to hold after at most steps and that on input with is guaranteed to hold after at most steps for polynomials and . Then the total length of the computation of counting also the computation steps of the oracle machine is bounded by a polynomial with . To carry this upper bound over to every branch of the computation tree of on input we have to solve a subtle problem: The upper bound on the execution length of on input applies to every computation path where for every oracle call , a correct computation path of on input is simulated. However, in our simulation of with oracle by a nondeterministic computation of on input , we possibly produce computation paths which on input can never reach, e.g.: if the correct answer of on oracle input is “yes”, then the continuation of the simulation of on input on all computation paths where answer “no” on oracle input is assumed, can never be reached by the computation of (with oracle ) on input . To make sure that the polynomial upper bound applies to every computation path of , we have to extend TM by a counter such that outputs and halts if more than steps have been executed.
Logarithmically bounded output. Suppose that is an arbitrary problem in and that, therefore, TM only has logarithmically many oracle calls. By similar considerations as for the polynomial time bound of the computation of on , we can make sure that the size of the output on every computation path of on is logarithmically bounded. To this end, we add a counter also for the number of oracle calls. The logarithmic bound (say for constants and ) on the size of the output applies to every computation path, where for every oracle call , a correct computation path of on input is simulated by . By adding to a counter for the size of the output, we can modify in such a way that outputs 0 and stops if the number of output bits of (which corresponds to the number of oracle calls of ) is exceeded. ∎
The above problems will allow us now to prove the  and completeness of the problems LexMaxSat and LogLexMaxSat, respectively. As mentioned above, credit for these results (in particular, the completeness of LexMaxSat) goes to Mark W. Krentel [15]. However, we hope that sticking to the standard terminology of oracle Turing machines (and avoiding the “detour” via the and classes based on a new machine model) will help to better convey the intuition of these results.
The LexMaxSat problem is complete and the LogLexMaxSat problem is complete. The hardness results hold even if is in 3CNF.
Proof.
The membership of LexMaxSat and the membership of LogLexMaxSat is seen by the following algorithm, which runs in deterministic polynomial time and has access to an oracle. Let denote the number of variables in (in case of the LexMaxSat problem) or the number of variables for which an order is given (in case of the LogLexMaxSat problem). The algorithm maintains a bit vector of the lexicographically maximal (prefix of a possible) model of . To this end, we initialize to and ask the following kind of questions to an oracle: Does there exist a model of such that the truth values of the first variables are and variable is set to 1? If the answer to this oracle call is “yes”, then we set ; otherwise we set . In either case, we then increment by 1 and continue with the next oracle call.
When is reached, then the algorithm checks the value of and stops with acceptance if and it stops with rejection if holds. If the number of variables of interest (i.e., those for which an order is given) is logarithmically bounded in the size of , then also the number of oracle calls of our algorithm is logarithmically bounded in the size of .
The hardness proof is by reduction from NPMax (resp. LogNPMax) to the LexMaxSat (resp. LogLexMaxSat) problem. Let be an arbitrary instance of NPMax (resp. LogNPMax). We make the following assumptions on the Turing machine : let have two tapes, where tape 1 serves as input and worktape while tape 2 is the dedicated output tape. This means that tape 2 initially has the starting symbol in cell 0 and blanks in all further tape cells. At time instant 0, the cursor on tape 2 makes a move to the right and leaves unchanged. At all later time instants, either the current symbol (i.e., blank ) and the cursor on tape 2 are both left unchanged or the current symbol is overwritten by 0 or 1 and the cursor is moved to the right. Recall that as output alphabet we have .
Following the standard proof of the CookLevin Theorem (see e.g. [21]), one can construct in polynomial time a propositional formula such that there is a onetoone correspondence between the computation paths of the nondeterministic TM on input and the satisfying truth assignments of . Let denote the maximum length of a computation path of on any input of length . Then formula is built from the following collection of variables:
 :

for , , , and , to express that at time instant of the computation, cell number of tape contains symbol , where denotes the set of all tape symbols of TM .
 :

for , , and , to express that at time instant , the cursor on tape points to cell number .
 :

for and , to express that at time instant , the NTM is in state , where denotes the set of all states of TM .
Suppose that denotes the maximum length of output strings of on any input of length . Then we introduce additional propositional variables and and we construct the propositional formula , where is defined as follows:
In other words, makes sure that the variables in encode the output string and encodes the last bit of the output string, i.e.: for every , variable is true in a model of if the th bit of the output string (along the computation path of corresponding to when considered as a model of ) is 1. Consequently, truth value false of can mean that, on termination of , the symbol in the th cell of tape 2 is either 0 or . The latter two cases are distinguished by the truth value of variable , which is false if and only if the symbol in the th cell of tape 2 is . Variable is true in a model of if the last bit of the output string is 1. The last bit (in the third line of the above definition of formula , this is position ) is recognized by the fact that beyond it, all cells on tape 2 contain the blank symbol. Note that the truth value of in any model of is uniquely determined by the truth values of the variables.
Finally, we transform into in 3CNF by some standard transformation (e.g., by the Tseytin transformation [26]). We thus introduce additional propositional variables such that every model of can be extended to a model of and every model of is also a model of . Now let denote the vector of variables , , and plus the additional variables introduced by the transformation into 3CNF. Let the variables in be arranged in arbitrary order and let denote the number of variables in .
In the reduction from NPMax to LexMaxSat, we define the following order on all variables in : . In the reduction from LogNPMax to LogLexMaxSat, we define an order only on part of the variables in , namely: (i.e., we ignore the variables in ).
We now construct a particular model of for which we will then show that it is in fact the lexicographically maximal model of . Let denote the lexicographically maximal output string produced by on input . Every computation path of corresponds to one or more models of . Consider the truth assignment on the variables , , and obtained as follows: restricted to is chosen such that it is lexicographically maximal among all truth assignments on corresponding to a computation path of on input and with output . Note that the truth assignments of on and are uniquely defined by the output of (i.e., no matter which concrete truth assignment on to encode a computation path of with this output we choose).
We claim that is the lexicographically maximal model of . To prove this claim, suppose to the contrary that there exists a lexicographically bigger model of . Then we distinguish 3 cases according to the group of variables where is bigger than :
(1) If is bigger than on , then (since the truth values of encode the output string of some computation of on ) there exists a bigger output than . This contradicts our assumption that is maximal. (2) If coincides with on and is bigger than on , then restricted to corresponds to a computation path producing the same output string as the computation path encoded by on . This contradicts our choice of truth assignment on . (3) The truth value of in any model of is uniquely determined by the truth value of . Hence, it cannot happen that and coincide on but differ on .
From the correspondence between the lexicographically maximal output string of on input and the lexicographically maximal model of , the correctness of our problem reductions (both, from NPMax to LexMaxSat and from LogNPMax to LogLexMaxSat) follows immediately, namely: For the LexMaxSat problem: the last bit in the lexicographically maximal output string of on input is 1 if and only if the truth value of variable in the lexicographically maximal model of is 1 (i.e., true). Likewise, for the LogLexMaxSat problem: the last bit in the lexicographically maximal output string of on input is 1 if and only if the truth value of variable in the lexicographically maximal truth assignment on that can be extended to a model of is 1 (i.e., true). Note that in case of the LogLexMaxSat problem, we may indeed simply ignore the variables because the truth values of and are uniquely determined by the output of – independently of the concrete choice of truth assignments to the variables. ∎
4. complete variants of SAT
In this section, we study two natural variants of Sat:

CardMinSat

Propositional formula and an atom .

Is true in a cardinalityminimal model of ?

CardMaxSat

Propositional formula and an atom .

Is true in a cardinalitymaximal model of ?
Both problems will be shown complete. We start with hardness for CardMaxSat.
CardMaxSat is hard, even for formulas in 3CNF.
Proof.
The hardness of CardMaxSat is shown by reduction from LogLexMaxSat. Consider an arbitrary instance of LogLexMaxSat, which is given by a propositional formula and an order over logarithmically many variables from . From this we construct an instance of CardMaxSat, which will be defined by a propositional formula and a dedicated variable in , namely . We simulate the lexicographical order over the variables by adding “copies” of each variable , i.e., for every , we introduce new variables with . The formula is now obtained from by adding the following subformulas. We add to the conjuncts . Hence, setting to true in a model of the resulting formula forces us to set all its “copies” to true. Finally, for the remaining variables in , we add “copies” and further extend by the conjuncts to make the cardinality of models indistinguishable on these variables.
Since , this transformation of into is feasible in polynomial time. Also note that is in 3CNF, whenever is in 3CNF. We claim that our problem reduction is correct, i.e., let denote the lexicographically maximal vector that can be extended to a model of . We claim that is true in a model of , s.t. is an extension of iff is true in a cardinalitymaximal model of . In fact, even a slightly stronger result can be shown, namely: if a model of is an extension of , then can be further extended to a cardinality maximal model of . Conversely, if is a cardinality maximal model of , then is also a model of and extends . ∎
Our ultimate goal in this section is to show that the completeness of CardMaxSat and also of CardMinSat hold even if restricted to the Krom case. In a first step, we reduce the CardMaxSat problem to a variant of the Independent Set problem, which we define next. Note that the standard reduction from 3SAT to Independent Set [21] is not sufficient: suppose we are starting off with a propositional formula with clauses. Then, if is satisfiable, every maximum independent set selects precisely vertices. Hence, additional work is needed to preserve the information on the cardinality of the models of . The variant of the Independent Set problem we require here is as follows: For the sake of readability, we first explicitly introduce a lower bound on the independent sets of interest and then we show how to encode this lower bound implicitly.

MaxIndependentSet

Undirected graph , vertex , and positive integer .

Is in a maximum independent set of with ?
Note that the question is not whether belongs to some independent set of size at least , but whether it belongs to a cardinalitymaximum independent set and whether this maximum value is . Without the restriction to a cardinalitymaximum independent set, the problem would obviously be in . With this restriction, the complexity increases, as we show next.
The MaxIndependentSet problem is hard.
Proof.
We extend the standard reduction from 3SAT to Independent Set [21] to a reduction from CardMaxSat to MaxIndependentSet. The result then follows from Lemma 4.
Let denote an arbitrary instance of CardMaxSat with , s.t. each clause is of the form where each is a literal. Let denote the set of variables in . We construct an instance of MaxIndependentSet where with “sufficiently large”, e.g., ; consists of vertices and ; and , for . It remains to specify the edges of :
(1) For every and every , contains edges , , and , i.e., contains triangles with .
(2) For every , s.t. and are complementary literals, contains edges with .
(3) For every and every , if is of the form , then contains edges ().
The intuition of this reduction is as follows: The triangles correspond to copies of the standard reduction from 3SAT to Independent Set [21]. Likewise, the edges between complementary literals are part of this standard reduction. It is easy to verify that for every independent set with , there also exists an independent set with , s.t. chooses from every copy of a triangle the “same” endpoint, i.e., for every , if and , then .
Since , the desired lower bound on the independent set can only be achieved if exactly one vertex is chosen from each triangle. We thus get the usual correspondence between models of and independent sets of of size . Note that this correspondence leaves some choice for those variables where the independent set contains no vertex corresponding to the literal . In such a case, we assume that the variable is set to true since in CardMaxSat, our goal is to maximize the number of variables set to true. Then a vertex may be added to an independent set with , only if no vertex corresponding to a literal of the form has been chosen into the independent set. Hence, if and only if is to true in the corresponding model of . ∎
CardMaxSat (resp. CardMinSat) is complete even when formulas are restricted to Krom and moreover the clauses consist of negative (resp. positive) literals only.
Proof.
Membership proceeds by the classical binary search [21] for finding the optimum, asking questions like “Does there exist a model of size at least ?” or “Does there exist a model of size at most ?”. With logarithmically many such calls to an oracle, the maximal (resp. minimal) size of all models of can thus be computed, and we check in a final question to an oracle if is true in a model of size .
For hardness, we start with the case of CardMaxSat. To this end, we first reduce MaxIndependentSet to the following intermediate problem: Given an undirected graph and vertex , is in a maximum independent set of ? The fact that this intermediate problem reduces to CardMaxSat then follows by expressing this problem via a Krom formula with propositional variables and clauses for every edge in . Hence, we obtain hardness also for the case of Krom formulas with negative literals only.
Hence, let us turn to the first reduction and consider an arbitrary instance of MaxIndependentSet with and . We define the corresponding instance of the intermediate problem with where with for fresh vertices , and . The additional edges in make sure that an independent set of contains either only vertices from or only vertices from . Clearly, contains the independent set with . This shows hardness of CardMaxSat.
To show the hardness result for CardMinSat restricted to Krom and positive literals, we now can give a reduction from CardMaxSat (restricted to Krom and negative literals). Given, such a CardMaxSat instance we construct where is given as
The intuition of variables is to represent that is assigned to false. We have the following observations: (1) a cardinalityminimal model of either sets or jointly, and , to true (for each variable ); i.e. it is of the form for some ; (2) for each , it holds that is a model of iff is a model of ; (3) for each , iff . It follows that is a yesinstance of CardMinSat iff is a yesinstance of CardMaxSat. ∎
5. Applications: Belief Revision and Abduction
In this section, we make use of the hardness results for CardMinSat in the previous section, in order to show novel complexity results for problems from the field of knowledge representation when restricted to the Krom fragment and the combined HornKrom fragment (i.e. Krom formulas with at most one positive literal per clause).
Belief Revision.
Belief revision aims at incorporating a new belief, while changing as little as possible of the original beliefs. We assume that a belief set is given by a propositional formula and that the new belief is given by a propositional formula . Revising by amounts to restricting the set of models of to those models which are “closest” to the models of . Several revision operators have been proposed which differ in how they define what “closest” means. Here, we focus on the revision operators due to Dalal [7] and Satoh [23].
Dalal’s operator measures the distance between the models of and in terms of the cardinality of model change, i.e., let and be two interpretations and let denote the symmetric difference between and , i.e., . Further, let denote the minimum number of propositional variables on which the models of and differ. We define , where denotes the models of a formula. Dalal’s operator is now given as: .
Satoh’s operator interprets the minimal change in terms of set inclusion. Thus let , where the operator applied to a set of sets selects those elements , s.t. contains no proper subset of . Then we define Satoh’s operator as .
Let denote a revision operator with . We analyse two wellstudied problems in belief revision.

BRImplication

Propositional formulas and , and an atom .

Does hold?

BRModel Checking

Propositional formulas and , and model of .

Does hold?
The complexity of these problems has been intensively studied [8, 17]. For arbitrary propositional formulas and , both problems are complete for Dalal’s revision operator. For Satoh’s revision operator, BRModel Checking (resp. BRImplication) is complete (resp. complete). Both [8] and [17] have also investigated the complexity of some restricted cases (e.g., when the formulas are restricted to Horn). Below, we pinpoint the complexity of these problems in the Krom case. Our hardness results will subsume known hardness results for the Horn case as well.
BRImplication and BRModel Checking are complete for Dalal’s revision operator even if the formulas and are restricted to both Krom and Horn form.
Proof.
The membership even holds without any restriction [8, 17]. Hardness is shown as follows: Given an instance of CardMinSat where is the set of variables in . By Theorem 4, this problem is complete even if is in Krom form with positive literals only. We define the following instances of BRImplication and BRModel Checking:
Let denote the set of clauses in . Let be a set of variables and let be two further variables. We define and as:
Obviously, both and are from the HornKrom fragment and can be constructed efficiently from .
Note that has exactly two models and . Let denote the size of a minimum model of . We claim that is contained in a minimum model of iff iff . The second equivalence is obvious. We prove both directions of the first equivalence separately.
First, suppose that has a minimum model with . Then the minimum distance of from models of is , which is witnessed by the model of (note that ). It can be easily verified that does not have smaller distance from any model of . At best, also has distance , namely from any model of of the form , s.t. is a minimum model of .
Now suppose that every model of with has size . Then the distance of from any model of is , since contains and at least elements from . On the other hand, the distance of from models of is witnessed by any model of where is a minimum model of .
In summary, we have thus shown that is guaranteed to hold. But holds if and only if there exists a minimum model of with . ∎
Compared to the hardness proof for Dalal’s revision operator in the Horn case [8, 17], our construction is much simpler, thanks to the previously established hardness result for CardMinSat. Moreover, it is not clear how the constructions from [8, 17] can be adapted to the HornKrom case.
The above theorem states that, for Dalal’s revision operator, the complexity of the BRImplication and BRModel Checking problem does not decrease even if the formulas are restricted to Krom and Horn form. In contrast, we shall show below that for Satoh’s revision, the complexity of BRImplication and BRModel Checking drops one level in the polynomial hierarchy if and are Krom. Hence, below, both the membership in case of Krom form and the hardness (which even holds for the restriction to Horn and Krom form) need to be proved. Also here, our hardness reductions differ substantially from those in [8, 17].
BRImplication is complete and BRModel Checking is complete for Satoh’s operator if the formulas and are restricted to Krom. Hardness holds even if and are further restricted to both Krom and Horn.
Proof.
For the membership proofs, recall the membership and membership proof of BRImplication and BRModel Checking for the Horn case in [8], Theorem 7.2; resp. [17], Theorem 20. The key idea there is that, given a model of and a model of the subsetminimality of can be tested in polynomial time by reducing this problem to a SAT problem involving the formulas and . The same idea holds for Krom form.
The crucial observation there is that, given a model of and a model of , one can check efficiently whether there exists a model of and a model of with . Indeed, let and let be fresh, pairwise distinct variables. Then a model of and a model of with exist iff for some variable , the following propositional formula is satisfiable:
(1) 
Here, denotes the formula that we obtain from by replacing every by . If and are Horn (and, likewise, if they are Krom), then (1) is Horn (resp. Krom) as well, and hence this satisfiability check is feasible in polynomial time.
Hardness is shown by the following reduction from 3SAT respectively co3SAT: Let be an arbitrary Boolean formula in 3CNF over the variables , i.e., , s.t. each clause is of the form , where the ’s are literals over . Let , , , and be sets of fresh, pairwise distinct propositional variables. We define and as follows:
where we set if for some and if . Finally, we define . Both and are from the HornKrom fragment and can be constructed efficiently from . We claim that is satisfiable iff iff .
Observe that every model of must set the variables in to true. The truth value of the variables in may be chosen arbitrarily. However, as soon as at least one is set to true, we must set to true due to the last conjunct in . Hence, is the only model of where is set to false. From this, the second equivalence above follows. Below we sketch the proof of the first equivalence.
First, suppose that is satisfiable and let be a model of . We set . Indeed, is a model of . We claim that is a witness for , i.e., is minimal. To prove this claim, let be a model of and a model of , s.t. . It suffices to show that then holds. To this end, we compare and first on then on and finally on : By the first group of conjuncts in , we may conclude from that and coincide on . In particular, both and are models of . But then they also coincide on since the second group of conjuncts in enforces false for every . Note that . Hence, no matter how we choose and on , we have .
Now suppose that is unsatisfiable. Let be a model of . To prove , we show that cannot be minimal. By the unsatisfiability of , we know that (restricted to ) is not a model of . Hence, there exists at least one clause which is false in . The proof goes by constructing with . The crucial observation is that the symmetric difference can be decreased by setting true and true. ∎
Abduction.
Abduction is used to produce explanations for some observed manifestations. Therefore, one of its primary fields of application is diagnosis. A propositional abduction problem (PAP) consists of a tuple , where is a finite set of variables, is the set of hypotheses, is the set of manifestations, and is a consistent theory in the form of a propositional formula. A set is a solution (also called explanation) to if is consistent and holds. A system diagnosis problem can be represented by a PAP as follows. The theory is the system description. The hypotheses describe the possibly faulty system components. The manifestations are the observed symptoms, describing the malfunction of the system. The solutions of are the possible explanations of the malfunction.
Often, one is not interested in any solution of a given PAP but only in minimal solutions, where minimality is defined w.r.t. some preorder on the powerset . Two natural preorders are setinclusion and smaller cardinality denoted as . Note that allowing any solution corresponds to choosing “” as the preorder. In [6] a trichotomy (of completeness, completeness, and tractability) has been proved for the Solvability problem of propositional abduction, i.e., deciding if a PAP has at least one solution (the preorder “” has thus been considered). Nordh and Zanuttini [19] have identified many further restrictions which make the Solvability problem tractable. All of the above mentioned preorders have been systematically investigated in [9]. A study of the counting complexity of abduction with these preorders has been carried out by Hermann and Pichler [13]. Of course, if a PAP has any solution than it also has a minimal solution. Hence, the preorder is only of interest for problems like the following one:

Relevance

PAP and hypothesis .

Is relevant, i.e., does have a minimal solution with ?
Known results [9] are as follows: The Relevance problem is complete for preorders and and complete for preorder . Moreover, the complexity drops by one level in the polynomial hierarchy if the theory is restricted to Horn. In [6], the Krom case was considered for the preorder . For the preorder , the Krom case was implicitly settled in [9]. Indeed, an inspection of the hardness proof of the Relevance problem in the Horn case reveals that hardness holds even if the theory is simultaneously restricted to Horn
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