Theory of Computing Systems

, Volume 53, Issue 3, pp 467–502

The Complexity of Computing Minimal Unidirectional Covering Sets

  • Dorothea Baumeister
  • Felix Brandt
  • Felix Fischer
  • Jan Hoffmann
  • Jörg Rothe
Article

DOI: 10.1007/s00224-012-9437-9

Cite this article as:
Baumeister, D., Brandt, F., Fischer, F. et al. Theory Comput Syst (2013) 53: 467. doi:10.1007/s00224-012-9437-9

Abstract

A common thread in the social sciences is to identify sets of alternatives that satisfy certain notions of stability according to some binary dominance relation. Examples can be found in areas as diverse as voting theory, game theory, and argumentation theory. Brandt and Fischer (in Math. Soc. Sci. 56(2):254–268, 2008) proved that it is NP-hard to decide whether an alternative is contained in some inclusion-minimal unidirectional (i.e., either upward or downward) covering set. For both problems, we raise this lower bound to the \(\varTheta_{2}^{p}\) level of the polynomial hierarchy and provide a \(\varSigma_{2}^{p}\) upper bound. Relatedly, we show that a variety of other natural problems regarding minimal or minimum-size unidirectional covering sets are hard or complete for either of NP, coNP, and \(\varTheta_{2}^{p}\). An important consequence of our results is that neither minimal upward nor minimal downward covering sets (even when guaranteed to exist) can be computed in polynomial time unless P=NP. This sharply contrasts with Brandt and Fischer’s result that minimal bidirectional covering sets are polynomial-time computable.

Keywords

Computational social choice Computational complexity Minimal upward covering sets Minimal downward covering sets 

1 Introduction

A common thread in the social sciences is to identify sets of alternatives that satisfy certain notions of stability according to some binary dominance relation. Applications range from cooperative to non-cooperative game theory, from social choice theory to argumentation theory, and from multi-criteria decision analysis to sports tournaments (see, e.g., [7, 33] and the references therein). To give an example from cooperative game theory, von Neumann and Morgenstern [40] introduced the notion of stable set as the set of (“efficient” and “individually rational”1) payoff vectors in a cooperative game that satisfies both internal stability (no vector in this set is dominated by another vector in the set) and external stability (every vector outside this set is dominated by some vector inside the set). The underlying dominance relation is defined as follows: A payoff vector \(\vec{x}=(x_{1},x_{2},\ldots,x_{n})\)dominates a payoff vector \(\vec{y}=(y_{1},y_{2},\ldots,y_{n})\) if there is a nonempty coalition C of players such that xi>yi for all iC and ∑iCxi is bounded above by the profit the players in C can make on their own. Stable sets exist for some, but not for all cooperative games [34], and if they exist, they need not be unique [35]. Brandt and Fischer [7] proved that every stable set is a “minimal upward covering set” and thus contained in the “upward uncovered set” (these notions, which are central to the present paper, will be defined formally in Sect. 2).

In settings of social choice, the most common dominance relation is the pairwise majority relation, where an alternative x is said to dominate another alternative y (written xy) if the number of individuals preferring x to y exceeds the number of individuals preferring y to x. McGarvey [36] proved that every asymmetric dominance relation can be realized via a particular preference profile, even if the individual preferences are linear.

For the set A={a,b,c,d} of alternatives, the dominance graph (A,≻) shown in Fig. 1 may for example result from the individual preferences of six voters given in the following table, where each column represents a number of voters with preferences given in decreasing order. For example, the first column represents two voters who rank the alternatives in alphabetical order. Observe that alternative a is preferred to alternative b by four out of six voters, which is why there is an edge from a to b (i.e., ab) in the corresponding dominance graph.

2

1

1

1

1

a

d

c

b

d

b

a

d

c

a

c

b

b

d

c

d

c

a

a

b

Fig. 1

Dominance graph (A,≻)

A well-known paradox due to the Marquis de Condorcet [13] says that the majority relation may contain cycles and thus does not always admit maximal elements, even if all of the underlying individual preferences do. Consider, for example, the three individual preference relations a>1b>1c, b>2c>2a, and c>3a>3b. Here, a majority prefers a to b and b to c, but also c to a. This means that although the individual preferences >i are each transitive, the resulting dominance relation (abca) is not, so the concept of maximality is rendered useless in such cases. For this reason, various alternative solution concepts that can be used in place of maximality for nontransitive relations (see, e.g., [33]) have been proposed. In particular, concepts based on covering relations—transitive subrelations of the dominance relation at hand—have turned out to be very attractive [16, 20, 39].

In this paper, we study the computational complexity of problems related to the notions of upward and downward covering sets in dominance graphs. An alternative x is said to upward cover another alternative y if x dominates y and every alternative dominating x also dominates y. The intuition is that x “strongly” dominates y in the sense that there is no alternative that dominates x but not y. Looking for example at the dominance graph (A,≻) in Fig. 1, although alternative a dominates alternative b, a does not upward cover b, since alternative d dominates a but not b. On the other hand, alternative b does upward cover alternative c, since b dominates c, and the only alternative dominating b, namely a, also dominates c.

Similarly, an alternative x is said to downward cover another alternative y if x dominates y and every alternative dominated by y is also dominated by x. The intuition here is that x “strongly” dominates y in the sense that there is no alternative dominated by y but not by x. Again looking at the dominance graph (A,≻) from Fig. 1, a downward covers b, since a dominates both b and c, the only alternative dominated by b. However, although b dominates c, b does not downward cover c, since b does not dominate d, which is dominated by c.

A minimal upward or minimal downward covering set is defined as an inclusion-minimal set of alternatives that satisfies certain notions of internal and external stability with respect to the upward or downward covering relation [7, 16] (cf. the von Neumann and Morgenstern stable sets in cooperative game theory mentioned in the first paragraph of the introduction), as will be formally stated in Definition 3 in Sect. 2.

Recent work in computational social choice has addressed the computational complexity of most solution concepts proposed in the context of binary dominance (see, e.g., [2, 7, 8, 9, 14, 53]). In particular, Brandt and Fischer [7] have shown that the minimal bidirectional covering set can be computed in polynomial time, where an alternative x is said to bidirectionally cover another alternative y if x covers y upward and downward. Due to its properties this set is particularly attractive from a social-choice-theoretic point of view (see the references cited in [7]). On the other hand, Brandt and Fischer [7] show NP-hardness of both the problem of deciding whether an alternative is contained in some minimal upward covering set and the problem of deciding whether an alternative is contained in some minimal downward covering set. For both problems, we improve on these results by raising their NP-hardness lower bounds to the \(\varTheta_{2}^{p}\) level of the polynomial hierarchy, and we provide an upper bound of \(\varSigma_{2}^{p}\). Moreover, we will analyze the complexity of a variety of other problems related to minimal and minimum-size upward and downward covering sets that have not been studied before. In particular, we provide hardness and completeness results for the complexity classes NP, coNP, and \(\varTheta_{2}^{p}\). A complete overview of our complexity results is presented in Tables 1 and 2 in Sect. 3. Remarkably, these new results imply that neither minimal upward covering sets nor minimal downward covering sets (even when guaranteed to exist) can be found in polynomial time unless P=NP. This sharply contrasts with Brandt and Fischer’s above-mentioned result that minimal bidirectional covering sets are polynomial-time computable [7]. Note that, notwithstanding the hardness of computing minimal upward covering sets, the decision version of this search problem is trivially in P: Every dominance graph always contains a minimal upward covering set.
Table 1

Overview of complexity results for the various types of upward covering set problems. As indicated, previously known results are due to Brandt and Fischer [7]; all other results are new to this paper

Problem type

\(\mbox {\textsc {MC}$_{\mathrm {u}}$}\)

\(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\)

Size

NP-complete, see Thm. 11

NP-complete, see Thm. 11

Member

\(\varTheta_{2}^{p}\)-hard and in \(\varSigma_{2}^{p}\), see Thm. 12

\(\varTheta_{2}^{p}\)-complete, see Thm. 13

Member-All

coNP-complete, see [7]

\(\varTheta_{2}^{p}\)-complete, see Thm. 13

Unique

coNP-hard and in \(\varSigma_{2}^{p}\), see Thm. 14

coNP-hard and in \(\varTheta_{2}^{p}\), see Thm. 16

Test

coNP-complete, see Thm. 14

coNP-complete, see Thm. 15

Find

not in polynomial time unless P=NP, see Thm. 17

not in polynomial time unless P=NP, see Thm. 17

Table 2

Overview of complexity results for the various types of downward covering set problems. As indicated, previously known results are due to Brandt and Fischer [7]; all other results are new to this paper

Problem type

\(\mbox {\textsc {MC}$_{\mathrm {d}}$}\)

\(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\)

Size

NP-complete, see Thm. 27

NP-complete, see Thm. 27

Member

\(\varTheta_{2}^{p}\)-hard and in \(\varSigma_{2}^{p}\), see Thm. 30

coNP-hard and in \(\varTheta_{2}^{p}\), see Thm. 28

Member-All

coNP-complete, see [7]

coNP-hard and in \(\varTheta_{2}^{p}\), see Thm. 28

Unique

coNP-hard and in \(\varSigma_{2}^{p}\), see Thm. 31

coNP-hard and in \(\varTheta_{2}^{p}\), see Thm. 28

Test

coNP-complete, see Thm. 31

coNP-complete, see Thm. 29

Find

not in polynomial time unless P=NP (follows from [7], see Thm. 32)

not in polynomial time unless P=NP, see Thm. 32

Put into a wider perspective, this work adds to a growing body of complexity and hardness results for the lower levels of the polynomial hierarchy, for problems arising in various areas, such as optimization problems in logic (see, e.g., [29, 49] and also the surveys by Schaefer and Umans [46, 47]), logic programming and reasoning (see, e.g., [17, 18] and also the survey by Eiter and Gottlob [19]), graph theory (see, e.g., [26, 27, 29, 51]), voting problems in social choice theory (see, e.g., [23, 28, 45] and also the survey by Hemaspaandra et al. [24]), and fair division problems in multiagent resource allocation (see, e.g., [6]).

This paper is organized as follows. Section 2 provides the needed definitions and notation, and Sect. 3 states all results and a discussion of the results. After presenting the constructions for minimal and minimum-size upward covering sets in Sect. 4.1, the proofs of the results on minimal and minimum-size upward covering sets are given in Sect. 4.2. Section 5.1 presents the constructions for minimal and minimum-size downward covering sets and Sect. 5.2 gives the proofs of the results on minimal and minimum-size downward covering sets. Finally Sect. 6 concludes this paper.

2 Definitions and Notation

In this section, we define the necessary concepts from social choice theory and complexity theory.

Definition 1

(Covering Relations)

Let A be a finite set of alternatives, let BA, and let ≻ ⊆A×A be a dominance relation on A, i.e., ≻ is asymmetric and irreflexive.2 A dominance relation ≻ on a set A of alternatives can be conveniently represented as a dominance graph, denoted by (A,≻), whose vertices are the alternatives from A, and for each x,yA there is a directed edge from x to y if and only if xy.

For any two alternatives x and y in B, define the following covering relations (see, e.g., [5, 20, 39]):
  • xupward coversyinB, denoted by \(x\, C_{u}^{B}\, y\), if xy and for all zB, zx implies zy, and

  • xdownward coversyinB, denoted by \(x\, C_{d}^{B}\, y\), if xy and for all zB, yz implies xz.

When clear from the context, we omit mentioning “in B” explicitly and simply write xCuy rather than \(x\, C_{u}^{B}\, y\), and xCdy rather than \(x\, C_{d}^{B}\, y\).

Definition 2

(Uncovered Set)

Let A be a set of alternatives, let BA be any subset, let ≻ be a dominance relation on A, and let C be a covering relation on A based on ≻. The uncovered set ofBwith respect toC is defined as
$$\mathrm{UC}_C(B)=\{y\in B \,\mid \:x\, C\, y \mbox{ for no } x \in B\}. $$
For notational convenience, let \(\mathrm{UC}_{z}(B) = \mathrm{UC}_{C_{z}}(B)\) for z∈{u,d}, and we call UCu(B) the upward uncovered set ofB and UCd(B) the downward uncovered set of B.

Example 1

(Upward and Downward Uncovered Set)

Since in the dominance graph (A,≻) from Fig. 1 in the introduction,
  • b upward covers c in A (i.e., \(b\, C_{u}^{A} \, c\)), but no element in A except c is upward covered, and

  • a downward covers b in A (i.e., \(a\, C_{d}^{A} \,b\)), but no element in A except b is downward covered,

UCu(A)={a,b,d} is the upward uncovered set and UCd(A)={a,c,d} is the downward uncovered set of A.

For both the upward and the downward covering relation (henceforth both will be called unidirectional covering relations), transitivity of the relation implies nonemptiness of the corresponding uncovered set for each nonempty set of alternatives. The intuition underlying covering sets is that there should be no reason to restrict the selection by excluding some alternative from it (internal stability) and there should be an argument against each proposal to include an outside alternative into the selection (external stability).

Definition 3

(Minimal Covering Set)

Let A be a set of alternatives, let ≻ be a dominance relation on A, and let C be a covering relation based on ≻. A subset BA is a covering set forAunderC if the following two properties hold:
  • Internal stability: UCC(B)=B.

  • External stability: For all yAB, \(y \not\in \mathrm{UC}_{C}(B \cup\{y\})\).

A covering set M for A under C is said to be (inclusion-)minimal if no M′⊂M is a covering set for A under C.

Example 2

(Minimal Upward and Downward Covering Set)

Again looking at the dominance graph (A,≻) from Fig. 1 in the introduction, note that A is neither an upward nor a downward covering set for itself, since internal stability is violated in both cases:
$$\mathrm{UC}_u(A) = \{a,b,d\} \neq A \neq\{a,c,d\} = \mathrm{UC}_d(A). $$
The set {a,b,d} is not an upward covering set for A either, again because it does not satisfy internal stability: UCu({a,b,d})={b,d}≠{a,b,d}, since d (being undominated in {a,b,d}) upward covers a. However, {b,d} is an upward covering set for A, because it satisfies both
  • internal stability, i.e., UCu({b,d})={b,d}, and

  • external stability, i.e., neither a∈UCu({a,b,d})={b,d} nor c∈UCu({b,c,d})={b,d}, the latter equality holding due to b (which is undominated in {b,c,d}) upward covering c.

Note that {b,d} is even a minimal upward covering set for A, since every strict subset of {b,d} violates external stability and thus is not an upward covering set for A. Moreover, {b,d} is the unique minimal upward covering set for A.

If the dominance relation ac were missing in (A,≻), then the resulting dominance graph would have two minimal upward covering sets for A, {a,c} and {b,d}. That is, minimal upward covering sets are not guaranteed to be unique.

The unique minimal downward covering set for A is {a,c,d}, since it satisfies both
  • internal stability, i.e., UCd({a,c,d})={a,c,d}, and

  • external stability, i.e., \(b \not\in\mathrm{UC}_{d}(A) = \{a,c,d\}\), as we have seen above,

and any strict subset of {a,c,d} is not a downward covering set for A, as can be easily verified.

Every upward uncovered set contains one or more minimal upward covering sets, whereas minimal downward covering sets may not always exist,3 and if they exist, they need not be unique [7]. Dutta [16] proposed minimal covering sets in the context of tournaments, i.e., complete dominance relations. In tournaments, both notions of covering coincide because the set of alternatives dominating a given alternative x consists precisely of those alternatives not dominated by x. Minimal unidirectional covering sets are one of several possible generalizations to incomplete dominance relations (for more details, see [7]). Occasionally, it might be helpful to specify the dominance relation explicitly to avoid ambiguity. In such cases we refer to the dominance graph used and write, e.g., “M is an upward covering set for (A,≻)”.

In addition to the (inclusion-)minimal unidirectional covering sets considered by Brandt and Fischer [7], we also consider minimum-size covering sets, i.e., unidirectional covering sets of smallest cardinality. Note that every minimum-size covering set is a minimal covering set; the converse, however, is not always true.4 For some of the computational problems we study, different complexities can be shown for the minimal and minimum-size versions of the problem (see Theorem 1 and Tables 1 and 2). Specifically, we consider six types of computational problems, for both upward and downward covering sets, and for each both their “minimal” (prefixed by MCu or MCd) and “minimum-size” (prefixed by MSCu or MSCd) versions. We first define the six problem types for the case of minimal upward covering sets:
  1. 1.

    \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\): Given a set A of alternatives, a dominance relation ≻ on A, and a positive integer k, does there exist some minimal upward covering set for A containing at most k alternatives?

     
  2. 2.

    \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\): Given a set A of alternatives, a dominance relation ≻ on A, and a distinguished element dA, is d contained in some minimal upward covering set for A?

     
  3. 3.

    \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\): Given a set A of alternatives, a dominance relation ≻ on A, and a distinguished element dA, is d contained in all minimal upward covering sets for A?

     
  4. 4.

    \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\): Given a set A of alternatives and a dominance relation ≻ on A, does there exist a unique minimal upward covering set for A?

     
  5. 5.

    \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\): Given a set A of alternatives, a dominance relation ≻ on A, and a subset MA, is M a minimal upward covering set for A?

     
  6. 6.

    \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\): Given a set A of alternatives and a dominance relation ≻ on A, find a minimal upward covering set for A.

     

If we replace “upward” by “downward” above, we obtain the six corresponding “downward covering” versions, denoted by \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\), \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\), \(\mbox{\textsc{MC}$_{\mathrm{d}}$}\hbox{-}\allowbreak\mbox{\textsc{Member}}\hbox{-}\allowbreak\mbox{\textsc{All}}\), \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\), \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\), and \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\). And if we replace “minimal” by “minimum-size” in the twelve problems just defined, we obtain the corresponding “minimum-size” versions: \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\), \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\), \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\), \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\), \({\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}}\), and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\).

Note that the four problems \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) are search problems, whereas the other twenty problems are decision problems.

We assume that the reader is familiar with the basic notions of complexity theory, such as polynomial-time many-one reducibility and the related notions of hardness and completeness, and also with standard complexity classes such as P, NP, coNP, and the polynomial hierarchy [38, 48] (see also, e.g., the textbooks [41, 44]). In particular, coNP is the class of sets whose complements are in NP. \(\varSigma_{2}^{p} = \mathrm {NP}^{ \mathrm {NP}}\), the second level of the polynomial hierarchy, consists of all sets that can be solved by an NP oracle machine that has access (in the sense of a Turing reduction) to an NP oracle set such as SAT. SAT denotes the satisfiability problem of propositional logic, which is one of the standard NP-complete problems (see, e.g., Garey and Johnson [21]) and is defined as follows: Given a boolean formula in conjunctive normal form, does there exist a truth assignment to its variables that satisfies the formula?

Papadimitriou and Zachos [43] introduced the class of problems solvable in polynomial time via asking \(\mathcal{O}(\log n)\) sequential Turing queries to NP. This class is also known as the \(\varTheta_{2}^{p}\) level of the polynomial hierarchy (see Wagner [52]), and has been shown to coincide with the class of problems that can be decided by a P machine that accesses its NP oracle in a parallel manner (see [22, 31]). Equivalently, \(\varTheta_{2}^{p}\) is the closure of NP under polynomial-time truth-table reductions. It follows immediately from the definitions that \(\mathrm {P}\subseteq \mathrm {NP}\cap \mathrm {coNP}\subseteq \mathrm {NP}\cup \mathrm {coNP}\subseteq \varTheta_{2}^{p} \subseteq\varSigma_{2}^{p}\).

\(\varTheta_{2}^{p}\) captures the complexity of various optimization problems. For example, the problem of testing whether the size of a maximum clique in a given graph is an odd number, the problem of deciding whether two given graphs have minimum vertex covers of the same size, and the problem of recognizing those graphs for which certain heuristics yield good approximations for the size of a maximum independent set or for the size of a minimum vertex cover each are known to be complete for \(\varTheta_{2}^{p}\) (see [26, 27, 51]). Hemaspaandra and Wechsung [29] proved that the minimization problem for boolean formulas is \(\varTheta_{2}^{p}\)-hard. In the field of computational social choice, the winner problems for Dodgson [15], Young [54], and Kemeny [30] elections have been shown to be \(\varTheta_{2}^{p}\)-complete in the nonunique-winner model [23, 28, 45], and also in the unique-winner model [25].

3 Results and Discussion

Results

Brandt and Fischer [7] proved that it is NP-hard to decide whether a given alternative is contained in some minimal unidirectional covering set. Using the notation of this paper, their results state that the problems \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) and \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) are NP-hard. The questions of whether these two problems are NP-complete or of higher complexity and whether minimal unidirectional covering sets can be found efficiently (when guaranteed to exist) were left open in [7]. Our contribution is
  1. 1.

    to raise Brandt and Fischer’s NP-hardness lower bounds for \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) and \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) to \(\varTheta_{2}^{p}\)-hardness and to provide (simple) \(\varSigma_{2}^{p}\) upper bounds for these problems, and

     
  2. 2.

    to extend the techniques we developed to apply also to the 22 other covering set problems defined in Sect. 2, in particular to the search problems.

     
Our results are stated in the following theorem.

Theorem 1

The complexity of the covering set problems defined in Sect2is as shown in Table 1for upward covering set problems and as shown in Table 2for downward covering set problems.

The detailed proofs of the single results collected in Theorem 1 will be presented in Sects. 4.2 for minimal and minimum-size upward covering sets and in Sect. 5.2 for minimal and minimum-size downward covering sets, and the technical constructions establishing the properties that are needed for these proofs are given in Sects. 4.1 for minimal and minimum-size upward covering sets and in Sect. 5.1 for minimal and minimum-size downward covering sets.

Discussion

We consider the problems of finding minimal and minimum-size upward and downward covering sets (\(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\), and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\)) to be particularly important and natural.

Regarding upward covering sets, we stress that our result (see Theorem 17) that, assuming P≠NP, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) are hard to compute does not seem to follow directly from the NP-hardness of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) in any obvious way. The decision version of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) is: Given a dominance graph, does it contain a minimal upward covering set? However, this question has always an affirmative answer, so the decision version of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) is trivially in P. Note also that \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) can be reduced in a “disjunctive truth-table” fashion to the search version of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) (“Given a dominance graph (A,≻) and an alternative dA, find some minimal upward covering set for A that contains d”) by asking this oracle set about all alternatives in parallel.5 So \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) is no harder (with respect to disjunctive truth-table reductions) than that problem. The converse, however, is not at all obvious. Brandt and Fischer’s results only imply the hardness of finding an alternative that is contained in all minimal upward covering sets [7]. Our reduction that raises the lower bound of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) from NP-hardness to \(\varTheta_{2}^{p}\)-hardness, however, also allows us to prove that \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) cannot be solved in polynomial time unless P=NP.

Regarding downward covering sets, the result that \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) cannot be computed in polynomial time unless P=NP is an immediate consequence of Brandt and Fischer’s result that it is NP-complete to decide whether there exists a minimal downward covering set [7, Thm. 9]. We provide an alternative proof based on our reduction showing that \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) is \(\varTheta_{2}^{p}\)-hard (see the proof of Theorem 32). In contrast to Brandt and Fischer’s proof, our proof shows that \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) is hard to compute even when the existence of a (minimal) downward covering set is guaranteed. As indicated in Tables 1 and 2, coNP-completeness of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\) and \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\) was also shown previously by Brandt and Fischer [7].

As mentioned above, the two problems \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) and \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) were already known to be NP-hard [7] and are here shown to be even \(\varTheta_{2}^{p}\)-hard. One may naturally wonder whether raising their (or any problem’s) lower bound from NP-hardness to \(\varTheta_{2}^{p}\)-hardness gives us any more insight into the problem’s inherent computational complexity. After all, P=NP if and only if \(\mathrm {P}= \varTheta_{2}^{p}\). However, this question is a bit more subtle than that and has been discussed carefully by Hemaspaandra et al. [24]. They make the case that the answer to this question crucially depends on what one considers to be the most natural computational model. In particular, they argue that raising NP-hardness to \(\varTheta_{2}^{p}\)-hardness potentially (i.e., unless longstanding open problems regarding the separation of the corresponding complexity classes could be solved) is an improvement in terms of randomized polynomial time (i.e., for the class RP introduced by Adleman [1]) and in terms of unambiguous polynomial time (i.e., for the class UP introduced by Valiant [50]): Since it is neither known whether NP=RP implies \(\varTheta_{2}^{p} = \mathrm {RP}\) nor whether NP=UP implies \(\varTheta_{2}^{p} = \mathrm {UP}\), proving \(\varTheta_{2}^{p}\)-hardness for the problems considered in this paper potentially gives a higher level of evidence (than merely NP-hardness) that these problems are neither in RP nor in UP [24].

4 Minimal and Minimum-Size Upward Covering Sets

In this section, we consider minimal and minimum-size upward covering sets.

4.1 Constructions

We start by giving the constructions that will be used for establishing results on the minimal and minimum-size upward covering set problems. Brandt and Fischer [7] proved the following result. Since we need their reduction in Construction 7 and Sect. 4.2, we give a proof sketch for Theorem 2.

Theorem 2

(Brandt and Fischer [7])

Deciding whether a designated alternative is contained in some minimal upward covering set for a given dominance graph is NP-hard. That is, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)is NP-hard.

Proof Sketch

NP-hardness is shown by a reduction from SAT. Given a boolean formula in conjunctive normal form, φ(v1,v2,…,vn)=c1c2∧…∧cr, over the set V={v1,v2,…,vn} of variables, construct an instance (A,≻,d) of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) as follows. The set of alternatives is
$$A=\bigl\{x_i,\overline{x}_i,x_i', \overline{x}_i' \,\mid \:v_i \in V\bigr\} \cup\{y_j \,\mid \:c_j \mbox{ is a clause in } \varphi\} \cup \{d\}, $$
where d is the distinguished alternative whose membership in some minimal upward covering set for A is to be decided, and the dominance relation ≻ is defined by:
  • For each i, 1≤in, there is a cycle \(x_{i} \succ \overline{x}_{i} \succ x_{i}' \succ\overline{x}_{i}' \succ x_{i}\);

  • if variable vi occurs in clause cj as a positive literal, then xiyj;

  • if variable vi occurs in clause cj as a negative literal, then \(\overline{x}_{i} \succ y_{j}\); and

  • for each j, 1≤jr, we have yjd.

As an example of this reduction, Fig. 2 shows the dominance graph resulting from the formula
$$(v_1 \vee\neg v_2 \vee v_3) \wedge(\neg v_1 \vee\neg v_3), $$
which is satisfiable, for example via the truth assignment that sets each of v1, v2, and v3 to false. Note that in this case the set \(\{\overline{x}_{1}, \overline{x}'_{1}, \overline{x}_{2}, \overline{x}'_{2}, \overline{x}_{3}, \overline{x}'_{3}\} \cup\{d\}\) is a minimal upward covering set for A corresponding to the satisfying assignment, so there indeed exists a minimal upward covering set for A that contains the designated alternative d. In general, Brandt and Fischer [7] proved that there exists a satisfying assignment for φ if and only if d is contained in some minimal upward covering set for A.
Fig. 2

Dominance graph for Theorem 2, example for the formula (v1∨¬v2v3)∧(¬v1∨¬v3)

 □

As we will use this reduction to prove results for both \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) and some of the other problems stated in Sect. 2, we now analyze the minimal and minimum-size upward covering sets of the dominance graph constructed in the proof sketch of Theorem 2. Brandt and Fischer [7] showed that each minimal upward covering set for A contains exactly two of the four alternatives corresponding to any of the variables, i.e., either xi and \(x_{i}'\), or \(\overline{x_{i}}\) and \(\overline{x_{i}}'\), 1≤in. We now assume that if φ is not satisfiable then for each truth assignment to the variables of φ, at least two clauses are unsatisfied (which can be ensured, if needed, by adding two dummy variables). It is easy to see that every minimal upward covering set for A not containing alternative d must consist of at least 2n+2 alternatives, where 2n alternatives are from the variables and at least two are from the unsatisfied clauses. Also, every minimal upward covering set for A containing d consists of exactly 2n+1 alternatives, where again 2n alternatives are from the variables, none from the clauses and alternative d. Thus, φ is satisfiable if and only if every minimum-size upward covering set consists of 2n+1 alternatives. These minimum-size upward covering sets always include alternative d. We summarize these observations in the following proposition for later use.

Proposition 1

For the reduction from SAT to\(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)presented in the proof sketch of Theorem 2, it holds that:
  1. 1.

    Every minimal upward covering set forAcontaining alternativedconsists of exactly 2n+1 alternatives.

     
  2. 2.

    Every minimal upward covering set forAnot containing alternativedmust consist of at least 2n+2 alternatives.

     
  3. 3.

    φis satisfiable if and only if every minimum-size upward covering set consists of 2n+1 alternatives (including d).

     

We now provide another construction that transforms a given boolean formula into a dominance graph with quite different properties.

Construction 3

(for coNP-hardness of upward covering set problems)

Given a boolean formula in conjunctive normal form, φ(w1,w2,…,wk)=f1f2∧…∧f, over the setW={w1,w2,…,wk} of variables, we construct a set of alternativesAand a dominance relationon A. Without loss of generality, we may assume that ifφis satisfiable then it has at least two satisfying assignments. This can be ensured, if needed, by adding dummy variables.

The set of alternatives is\(A=\{u_{i},\overline{u}_{i},u_{i}',\overline{u}_{i}' \,\mid \:w_{i} \in W\} \cup\{e_{j}, e_{j}' \,\mid \:f_{j}\ \mathrm{is}\ \mathrm{a}\ \mathrm{clause} \mathrm{in}\ \varphi\} \cup\{a_{1},a_{2},a_{3}\}\), and the dominance relationis defined by:
  • For each i, 1≤ik, there is a cycle\(u_{i} \succ \overline{u}_{i} \succ u_{i}' \succ\overline{u}_{i}' \succ u_{i}\);

  • if variablewioccurs in clausefjas a positive literal, thenuiej, \(u_{i} \succ e'_{j}\), \(e_{j} \succ\overline{u}_{i}\), and\(e'_{j} \succ\overline{u}_{i}\);

  • if variablewioccurs in clausefjas a negative literal, then\(\overline{u}_{i} \succ e_{j}\), \(\overline{u}_{i} \succ e'_{j}\), ejui, and\(e'_{j} \succ u_{i}\);

  • if variablewidoes not occur in clause fj, then\(e_{j} \succ u_{i}'\)and\(e_{j}' \succ\overline{u}_{i}'\);

  • for eachj, 1≤j, we havea1ejand\(a_{1} \succ e'_{j}\); and

  • there is a cyclea1a2a3a1.

Figure 3 shows some parts of the dominance graph that results from the given boolean formula φ. In particular, Fig. 3(a) shows that part of this graph that corresponds to some variable wi occurring in clause fj as a positive literal; Fig. 3(b) shows that part of this graph that corresponds to some variable wi occurring in clause fj as a negative literal; and Fig. 3(c) shows that part of this graph that corresponds to some variable wi not occurring in clause fj.
Fig. 3

Parts of the dominance graph defined in Construction 3

As a more complete example, Fig. 4 shows the entire dominance graph that corresponds to the concrete formula (¬w1w2)∧(w1∨¬w3), which can be satisfied by setting, for example, each of w1, w2, and w3 to true. A minimal upward covering set for A corresponding to this assignment is \(M=\{u_{1},u'_{1},u_{2},u'_{2},u_{3},u'_{3},a_{1},a_{2},a_{3}\}\). Note that neither e1 nor e2 occurs in M, and none of them occurs in any other minimal upward covering set for A either. For alternative e1 in the example shown in Fig. 4, this can be seen as follows. If there were a minimal upward covering set M′ for A containing e1 (and thus also \(e'_{1}\), since they both are dominated by the same alternatives) then neither \(\overline{u}_{1}\) nor u2 (which dominate e1) must upward cover e1 in M′, so all alternatives corresponding to the variables w1 and w2 (i.e., \(\{u_{i}, \overline{u}_{i}, u_{i}', \overline{u}_{i}' \,\mid \:i \in\{1,2\}\}\)) would also have to be contained in M′. Due to \(e_{1} \succ u'_{3}\) and \(e'_{1} \succ\overline{u}'_{3}\), all alternatives corresponding to w3 (i.e., \(\{u_{3}, \overline{u}_{3}, u_{3}', \overline{u}_{3}'\}\)) are in M′ as well. Note that, e2 and \(e'_{2}\) are no longer upward covered and must also be in M′. The alternatives a1,a2, and a3 are contained in every minimal upward covering set for A. But then M′ is not minimal because the upward covering set M, which corresponds to the satisfying assignment stated above, is a strict subset of M′. Hence, e1 cannot be contained in any minimal upward covering set for A.
Fig. 4

Dominance graph from Construction 3, example for the formula (¬w1w2)∧(w1∨¬w3)

We now show some properties of the dominance graph created by Construction 3 in general. We will need these properties for the proofs in Sect. 4.2. The first property, stated in Claim 4, has already been seen in the example above.

Claim 4

Consider the dominance graph (A,≻) created by Construction 3, and fix any j, 1≤j. For each minimal upward covering set M for A, if M contains the alternative ej then all other alternatives are contained in M as well (i.e., A=M).

Proof

To simplify notation, we prove the claim only for the case of j=1. However, since there is nothing special about e1 in our argument, the same property can be shown by an analogous argument for each j, 1≤j.

Let M be any minimal upward covering set for A, and suppose that e1M. First note that the alternatives dominating e1 and \(e_{1}'\) are always the same (albeit e1 and \(e_{1}'\) may dominate different alternatives). Thus, for each minimal upward covering set, either both e1 and \(e_{1}'\) are contained in it, or they both are not. Thus, since e1M, we have \(e_{1}' \in M\) as well.

Since the alternatives a1, a2, and a3 form an undominated three-cycle, they each are contained in every minimal upward covering set for A. In particular, {a1,a2,a3}⊆M. Furthermore, no alternative ej or \(e_{j}'\), 1≤j, can upward cover any other alternative in M, because a1M and a1 dominates ej and \(e_{j}'\) but none of the alternatives that are dominated by either ej or \(e_{j}'\). In particular, no alternative in any of the k four-cycles \(u_{i} \succ\overline{u}_{i} \succ u'_{i} \succ\overline{u}'_{i} \succ u_{i}\) can be upward covered by any alternative ej or \(e_{j}'\), and so they each must be upward covered within their cycle. For each of these cycles, every minimal upward covering set for A must contain at least one of the sets \(\{u_{i}, u'_{i}\}\) and \(\{\overline{u}_{i}, \overline{u}'_{i}\}\), since at least one is needed to upward cover the other one.6

Since e1M and by internal stability, we have that no alternative from M upward covers e1. In addition to a1, the alternatives dominating e1 are ui (for each i such that wi occurs as a positive literal in f1) and \(\overline{u}_{i}\) (for each i such that wi occurs as a negative literal in f1).

First assume that, for some i, wi occurs as a positive literal in f1. Suppose that \(\{u_{i}, u'_{i}\} \subseteq M\). If \(\overline{u}'_{i} \not\in M\) then e1 would be upward covered by ui, which is impossible. Thus \(\overline{u}'_{i} \in M\). But then \(\overline{u}_{i} \in M\) as well, since ui, the only alternative that could upward cover \(\overline{u}_{i}\), is itself dominated by \(\overline{u}'_{i}\). For the latter argument, recall that \(\overline{u}_{i}\) cannot be upward covered by any ej or \(e_{j}'\). Thus, we have shown that \(\{u_{i}, u'_{i}\} \subseteq M\) implies \(\{\overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\). Conversely, suppose that \(\{\overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\). Then \(u_{i}'\) is no longer upward covered by \(\overline{u}_{i}\) and hence must be in M as well. The same holds for the alternative ui, so \(\{\overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\) implies \(\{u_{i}, u'_{i}\} \subseteq M\). Summing up, if e1M then \(\{u_{i}, u'_{i}, \overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\) for each i such that wi occurs as a positive literal in f1.

By symmetry of the construction, an analogous argument shows that if e1M then \(\{u_{i}, u'_{i}, \overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\) for each i such that wi occurs as a negative literal in f1.

Now, consider any i such that wi does not occur in f1. We have \(e_{1} \succ u'_{i}\) and \(e'_{1} \succ\overline{u}'_{i}\). Again, none of the sets \(\{u_{i}, u'_{i}\}\) and \(\{\overline{u}_{i},\overline{u}'_{i}\}\) alone can be contained in M, since otherwise either ui or \(\overline{u}'_{i}\) would remain upward uncovered. Thus, e1M again implies that \(\{u_{i}, u'_{i}, \overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\).

Now it is easy to see that, since \(\bigcup_{1\leq i \leq k} \{u_{i}, u'_{i}, \overline{u}_{i}, \overline{u}'_{i}\} \subseteq M\) and since a1 cannot upward cover any of the ej and \(e_{j}'\), 1≤j, external stability of M enforces that \(\bigcup_{1 < j \leq\ell} \{e_{j}, e_{j}'\} \subseteq M\). Summing up, we have shown that if e1 is contained in any minimal upward covering set M for A, then M=A. □

Claim 5

Consider Construction 3. The boolean formula φ is satisfiable if and only if there is no minimal upward covering set for A that contains any of the ej, 1≤j.

Proof

It is enough to prove the claim for the case j=1, since the other cases can be proven analogously.

From left to right, suppose there is a satisfying assignment α:W→{0,1} for φ. Define the set
$$B_{\alpha} = \{a_1, a_2, a_3\} \cup \bigl\{u_i, u'_i \,\mid \:\alpha (w_i) = 1\bigr\} \cup\bigl\{\overline{u}_i, \overline{u}'_i \,\mid \:\alpha(w_i) = 0 \bigr\}. $$

Since every upward covering set for A must contain {a1,a2,a3} and at least one of the sets \(\{u_{i}, u'_{i}\}\) and \(\{\overline{u}_{i}, \overline{u}'_{i}\}\) for each i, 1≤ik, Bα is a (minimal) upward covering set for A. Let M be an arbitrary minimal upward covering set for A. By Claim 4, if e1 were contained in M, we would have M=A. But since BαA=M, this contradicts the minimality of M. Thus \(e_{1} \not\in M\).

From right to left, let M be an arbitrary minimal upward covering set for A and suppose \(e_{1} \not\in M\). By Claim 4, if any of the ej, 1<j, were contained in M, it would follow that e1M, a contradiction. Thus, {ej ∣ 1≤j}∩M=∅. It follows that each ej must be upward covered by some alternative in M. It is easy to see that for each j, 1≤j, and for each i, 1≤ik, ej is upward covered in \(M\cup\{e_{j}\}\supseteq\{u_{i},u'_{i}\}\) if wi occurs in fj as a positive literal, and ej is upward covered in \(M\cup\{e_{j}\}\supseteq\{\overline{u}_{i}, \overline{u}'_{i}\}\) if wi occurs in fj as a negative literal. It can never be the case that all four alternatives, \(\{u_{i}, u'_{i}, \overline{u}_{i}, \overline{u}'_{i}\}\), are contained in M, because then either ej would no longer be upward covered in M or the resulting set M was not minimal. Now, M induces a satisfying assignment for φ by setting, for each i, 1≤ik, α(wi)=1 if uiM, and α(wi)=0 if \(\overline{u}_{i} \in M\). □

Note that in Construction 3 every minimal upward covering set for A obtained from any satisfying assignment for φ contains exactly 2k+3 alternatives, and there is no minimal upward covering set of smaller size for A when φ is unsatisfiable.

Claim 6

Consider Construction 3. The boolean formula φ is not satisfiable if and only if there is a unique minimal upward covering set for A.

Proof

Recall that we assumed in Construction 3 that if φ is satisfiable then it has at least two satisfying assignments.

From left to right, suppose there is no satisfying assignment for φ. By Claim 5, there must be a minimal upward covering set for A containing one of the ej, 1≤j, and by Claim 4 this minimal upward covering set for A must contain all alternatives. By reason of minimality, there cannot be another minimal upward covering set for A.

From right to left, suppose there is a unique minimal upward covering set for A. Due to our assumption that if φ is satisfiable then there are at least two satisfying assignments, φ cannot be satisfiable, since if it were, there would be two distinct minimal upward covering sets corresponding to these assignments (as argued in the proof of Claim 5). □

Wagner provided a sufficient condition for proving \(\varTheta_{2}^{p}\)-hardness that was useful in various other contexts (see, e.g., [23, 26, 27, 29, 51]) and is stated here as Lemma 1.

Lemma 1

(Wagner [51])

LetSbe some NP-complete problem and letTbe any set. If there exists a polynomial-time computable functionfsuch that, for allm≥1 and all stringsx1,x2,…,x2msatisfying that ifxjSthenxj−1S, 1<j≤2m, we havethenTis\(\varTheta_{2}^{p}\)-hard.

We will apply Lemma 1 as well. In contrast with those previous results, however, one subtlety in our construction is due to the fact that we consider not only minimum-size but also (inclusion-)minimal covering sets. To the best of our knowledge, our Construction 7 and Construction 24, which will be presented later, for the first time apply Wagner’s technique [51] to problems defined in terms of minimality/maximality rather than minimum/maximum size of a solution:7 In Construction 7 below, we define a dominance graph based on Construction 3 and the construction presented in the proof sketch of Theorem 2 such that Lemma 1 can be applied to prove \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)\(\varTheta_{2}^{p}\)-hard (see Theorem 12), making use of the properties established in Claims 4, 5, and 6.

Construction 7

(for applying Lemma 1 to upward covering set problems)

We apply Wagner’s lemma with the NP-complete problemS=SAT and construct a dominance graph. Fix an arbitrarym≥1 and letφ1,φ2,…,φ2mbe 2mboolean formulas in conjunctive normal form such that ifφjis satisfiable then so isφj−1, for each j, 1<j≤2m. Without loss of generality, we assume that for each j, 1≤j≤2m, the first variable ofφjdoes not occur in all clauses ofφj. Furthermore, we requireφjto have at least two unsatisfied clauses ifφjis not satisfiable, and to have at least two satisfying assignments ifφjis satisfiable. It is easy to see that ifφjdoes not have this property, it can be transformed into a formula that does have it, without affecting the satisfiability of the formula.

We now define a polynomial-time computable function f, which maps the given 2mboolean formulas to a dominance graph (A,≻) with useful properties for upward covering sets. Define\(A = \bigcup_{j=1}^{2m} A_{j}\)and the dominance relationonAby
$$\Biggl(\bigcup_{j=1}^{2m} \succ_j \Biggr) \cup \Biggl(\bigcup_{i=1}^{m} \bigl\{\bigl(u_{1,2i}',d_{2i-1}\bigr), \bigl( \overline{u}_{1,2i}',d_{2i-1}\bigr) \bigr\} \Biggr) \cup \Biggl(\bigcup_{i=2}^{m} \bigl \{(d_{2i-1}, z) \,\mid \:z \in A_{2i-2} \bigr\} \Biggr), $$
where we use the following notation:
  1. (1)

    For each i, 1≤im, let (A2i−1,≻2i−1) be the dominance graph that results from the formulaφ2i−1according to Brandt and Fischer’s construction [7] given in the proof sketch of Theorem 2. We use the same names for the alternatives inA2i−1as in that proof sketch, except that we attach the subscript 2i−1. For example, alternativedfrom the proof sketch of Theorem 2 now becomesd2i−1, x1becomesx1,2i−1, y1becomesy1,2i−1, and so on.

     
  2. (2)

    For each i, 1≤im, let (A2i,≻2i) be the dominance graph that results from the formulaφ2iaccording to Construction 3. We use the same names for the alternatives inA2ias in that construction, except that we attach the subscript 2i. For example, alternativea1from Construction 3 now becomesa1,2i, e1becomese1,2i, u1becomesu1,2i, and so on.

     
  3. (3)

    For eachi, 1≤im, connect the dominance graphs (A2i−1,≻2i−1) and (A2i,≻2i) as follows. Let\(u_{1,2i}, \overline{u}_{1,2i}, u_{1,2i}', \overline{u}_{1,2i}' \in A_{2i}\)be the four alternatives in the cycle corresponding to the first variable ofφ2i. Then both\(u_{1,2i}'\)and\(\overline{u}_{1,2i}'\)dominated2i−1. The resulting dominance graph is denoted by\((B_{i}, \succ_{i}^{B})\).

     
  4. (4)

    Connect themdominance graphs\((B_{i}, \succ_{i}^{B})\), 1≤im, as follows: For eachi, 2≤im, d2i−1dominates all alternatives inA2i−2.

     
The dominance graph (A,≻) is sketched in Fig. 5. Clearly, (A,≻) is computable in polynomial time.
Fig. 5

Dominance graph from Construction 7. Most alternatives, and all edges between pairs of alternatives, in Aj, 1≤j≤2m, have been omitted. All edges between alternatives in Ai and alternatives in Aj for ij are shown. An edge incident to a set of alternatives represents an edge incident to each alternative in the set

Before we use this construction to obtain \(\varTheta_{2}^{p}\)-hardness results for some of our upward covering set problems in Sect. 4.2, we again show some useful properties of the dominance graph constructed, and we first consider the dominance graph \((B_{i}, \succ_{i}^{B})\) (see Step (3) in Construction 7) separately,8 for any fixed i with 1≤im. Doing so will simplify our argument for the whole dominance graph (A,≻). Recall that \((B_{i},\succ_{i}^{B})\) results from the formulas φ2i−1 and φ2i.

Claim 8

Consider Construction 7. Alternative d2i−1 is contained in some minimal upward covering set for \((B_{i},\succ_{i}^{B})\) if and only if φ2i−1 is satisfiable and φ2i is not satisfiable.

Proof

Distinguish the following three cases.
Case 1:

φ2i−1∈SAT and φ2i∈SAT. Since φ2i is satisfiable, it follows from the proof of Claim 5 that for each minimal upward covering set M for \((B_{i},\succ_{i}^{B})\), either \(\{u_{1,2i}, u'_{1,2i}\} \subseteq M\) or \(\{\overline{u}_{1,2i}, \overline{u}'_{1,2i}\} \subseteq M\), but not both, and that none of the ej,2i and \(e'_{j,2i}\) is in M. If \(\overline{u}'_{1,2i} \in M\) but \(u'_{1,2i} \not\in M\), then \(d_{2i-1} \not\in\mathrm{UC}_{u}(M)\), since \(\overline{u}'_{1,2i}\) upward covers d2i−1 within M. If \(u'_{1,2i} \in M\) but \(\overline{u}_{1,2i} \not\in M\), then \(d_{2i-1} \not\in\mathrm{UC}_{u}(M)\), since \(u'_{1,2i}\) upward covers d2i−1 within M. Hence, by internal stability, d2i−1 is not contained in M.

Case 2:

\(\varphi_{2i-1} \not\in \mathrm {SAT}\) and \(\varphi_{2i} \not\in \mathrm {SAT}\). Since \(\varphi_{2i-1} \not\in \mathrm {SAT}\), it follows from the construction used in the proof of Theorem 2 that each minimal upward covering set M for \((B_{i},\succ_{i}^{B})\) contains at least one alternative yj,2i−1 (corresponding to some clause of φ2i−1) that upward covers d2i−1. Thus d2i−1 cannot be in M, again by internal stability.

Case 3:

φ2i−1∈SAT and \(\varphi_{2i} \not\in \mathrm {SAT}\). Since φ2i−1∈SAT, it follows from the proof of Theorem 2 (see also Proposition 1) that there exists a minimal upward covering set M′ for (A2i−1,≻2i−1) that corresponds to a satisfying truth assignment for φ2i−1. In particular, none of the yj,2i−1 is in M′. On the other hand, since \(\varphi_{2i} \not\in \mathrm {SAT}\), it follows from Claim 6 that A2i is the only minimal upward covering set for (A2i,≻2i). Define M=M′∪A2i. It is easy to see that M is a minimal upward covering set for \((B_{i},\succ_{i}^{B})\), since the only edges between A2i−1 and A2i are those from \(\overline{u}'_{1,2i}\) and \(u'_{1,2i}\) to d2i−1, and both \(\overline{u}'_{1,2i}\) and \(u'_{1,2i}\) are dominated by elements in M not dominating d2i−1.

We now show that d2i−1M. Note that \(\overline{u}'_{1,2i}\), \(u'_{1,2i}\), and the yj,2i−1 are the only alternatives in Bi that dominate d2i−1. Since none of the yj,2i−1 is in M, they do not upward cover d2i−1. Also, \(u'_{1,2i}\) doesn’t upward cover d2i−1, since \(\overline{u}_{1,2i} \in M\) and \(\overline{u}_{1,2i}\) dominates \(u'_{1,2i}\) but not d2i−1. On the other hand, by our assumption that the first variable of φ2i does not occur in all clauses, there exist alternatives ej,2i and \(e'_{j,2i}\) in M that dominate \(\overline{u}'_{1,2i}\) but not d2i−1, so \(\overline{u}'_{1,2i}\) doesn’t upward cover d2i−1 either. Thus d2i−1M.

Note that, by our assumption on how the formulas are ordered, the fourth case (i.e., \(\varphi_{2i-1} \not\in \mathrm {SAT}\) and φ2i∈SAT) cannot occur. Thus, the proof is complete. □

Claim 9

Consider Construction 7. For each i, 1≤im, let Mi be a minimal upward covering set for \((B_{i},\succ_{i}^{B})\) according to the cases in the proof of Claim 8. Then each of the sets Mi must be contained in every minimal upward covering set for (A,≻).

Proof

The minimal upward covering set Mm for \((B_{m},\succ_{m}^{B})\) must be contained in every minimal upward covering set for (A,≻), since no alternative in ABm dominates any alternative in Bm. On the other hand, for each i, 1≤i<m, no alternative in Bi can be upward covered by d2i+1 (which is the only element in ABi that dominates any of the elements of Bi), since d2i+1 is dominated within every minimal upward covering set for Bi+1 (and, in particular, within Mi+1). Thus, each of the sets Mi, 1≤im, must be contained in every minimal upward covering set for (A,≻). □

Claim 10

Consider Construction 7. It holds that

Proof

To show (4.2) from left to right, suppose ∥{i ∣ φi∈SAT}∥ is odd. Recall that for each j, 1<j≤2m, if φj is satisfiable then so is φj−1. Thus, there exists some i, 1≤im, such that φ1,…,φ2i−1∈SAT and \(\varphi_{2i}, \dots, \varphi_{2m} \not\in \mathrm {SAT}\). In Case 3 in the proof of Claim 8 we have seen that there is some minimal upward covering set for \((B_{i},\succ_{i}^{B})\)—call it Mi—that corresponds to a satisfying assignment of φ2i−1 and that contains all alternatives of A2i. Note that, Mi contains d2i−1. For each ji, 1≤jm, let Mj be some minimal upward covering set for \((B_{j},\succ_{j}^{B})\) according to Case 1 (if j<i) and Case 2 (if j>i) in the proof of Claim 8.

In Case 1 in the proof of Claim 8 we have seen that d2i−3 is upward covered either by \(\overline{u}'_{1,2i-3}\) or by \(u'_{1,2i-3}\). This is no longer the case, since d2i−1 is in Mi and it dominates all alternatives in A2i−2 but not d2i−3. By assumption, φ2i−3 is satisfiable, so there exists a minimal upward covering set that contains d2i−3 as well. Let
$$M = \{d_1, d_3, \dots, d_{2i-1}\} \cup\bigcup _{1 \leq j \leq m} M_j . $$
By Claim 9, and by observing that all elements not in M are upward covered, it follows that M is a minimal upward covering set for (A,≻) that contains d1.

To show (4.2) from right to left, suppose that ∥{i ∣ φi∈SAT}∥ is even. For a contradiction, suppose that there exists some minimal upward covering set M for (A,≻) that contains d1. If \(\varphi_{1} \not\in \mathrm {SAT}\) then we immediately obtain a contradiction by the argument in the proof of Theorem 2. On the other hand, if φ1∈SAT then our assumption that ∥{i ∣ φi∈SAT}∥ is even implies that φ2∈SAT. It follows from the proof of Claim 4, and from Claim 9, that every minimal upward covering set for (A,≻) (thus, in particular, M) contains either \(\{u_{1,2i}, u'_{1,2i}\}\) or \(\{\overline{u}_{1,2i}, \overline{u}'_{1,2i}\}\), but not both, and that none of the ej,2i and \(e'_{j,2i}\) is in M. By the argument presented in Case 3 in the proof of Claim 8, the only way to prevent d1 from being upward covered by an element of M, either \(u_{1,2}'\) or \(\overline{u}_{1,2}'\), is to include d3 in M as well.9 By applying the same argument m−1 times, we will eventually reach a contradiction, since d2m−1M can no longer be prevented from being upward covered by an element of M, either \(u_{1,2m}'\) or \(\overline{u}_{1,2m}'\). Thus, no minimal upward covering set M for (A,≻) contains d1, which completes the proof of (4.2). □

Furthermore, it holds that ∥{i ∣ φi∈SAT}∥ is odd if and only if d1 is contained in all minimum-size upward covering sets for A. This is true since the minimal upward covering sets for A that contain d1 are those that correspond to some satisfying assignment for all satisfiable formulas φi, and as we have seen in the analysis of Construction 3 and the proof sketch of Theorem 2 (see also Proposition 1), these are the minimum-size upward covering sets for A.

4.2 Proofs

In this section, we prove the parts of Theorem 1 that consider minimal and minimum-size upward covering sets by applying the constructions and the properties of the resulting dominance graphs presented in Sect. 4.1.

Theorem 11

It is NP-complete to decide, given a dominance graph (A,≻) and a positive integer k, whether there exists a minimal/minimum-size upward covering set forAof size at most k. That is, both\(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\)and\(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\)are NP-complete.

Proof

This result can be proven by using the construction of Theorem 2. Let φ be a given boolean formula in conjunctive normal form, and let n be the number of variables occurring in φ. Setting the bound k for the size of a minimal/minimum-size upward covering set to 2n+1 proves that both problems are hard for NP. Indeed, as we have seen in the paragraph after the proof sketch of Theorem 2 (see also Proposition 1), there is a size 2n+1 minimal upward covering set (and hence a minimum-size upward covering set) for A if and only if φ is satisfiable. Both problems are NP-complete, since they can obviously be decided in nondeterministic polynomial time. □

Theorem 12

Deciding whether a designated alternative is contained in some minimal upward covering set for a given dominance graph is hard for\(\varTheta_{2}^{p}\)and in \(\varSigma_{2}^{p}\). That is, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)is hard for\(\varTheta_{2}^{p}\)and in \(\varSigma_{2}^{p}\).

Proof

\(\varTheta_{2}^{p}\)-hardness follows directly from Claim 10, which applies Wagner’s lemma to upward covering set problems. Specifically, this claim shows that in Construction 7 the alternative d1 is contained in some minimal upward covering set for A if and only if the number of underlying boolean formulas that are satisfiable is odd. For the upper bound, let (A,≻) be a dominance graph and d a designated alternative in A. First, observe that we can verify in polynomial time whether a subset of A is an upward covering set for A, simply by checking whether it satisfies internal and external stability. Now, we can guess an upward covering set BA with dB in nondeterministic polynomial time and verify its minimality by checking that none of its subsets is an upward covering set for A. This places the problem in NPcoNP and consequently in \(\varSigma_{2}^{p}\). □

Theorem 13

  1. 1.

    It is\(\varTheta_{2}^{p}\)-complete to decide whether a designated alternative is contained in some minimum-size upward covering set for a given dominance graph. That is, \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)is\(\varTheta_{2}^{p}\)-complete.

     
  2. 2.

    It is\(\varTheta_{2}^{p}\)-complete to decide whether a designated alternative is contained in all minimum-size upward covering sets for a given dominance graph. That is, \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\)is\(\varTheta_{2}^{p}\)-complete.

     

Proof

Wagner’s lemma can be used to show \(\varTheta_{2}^{p}\)-hardness for both problems. The remark made after Claim 10 says that in Construction 7 the alternative d1 is contained in all minimum-size upward covering sets for A if and only if the number of underlying boolean formulas that are satisfiable is odd. Hence \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\) are both \(\varTheta_{2}^{p}\)-hard.

To see that \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) is contained in \(\varTheta_{2}^{p}\), let (A,≻) be a dominance graph and d a designated alternative in A. Obviously, in nondeterministic polynomial time we can decide, given (A,≻), xA, and some positive integer ≤∥A∥, whether there exists some upward covering set B for A such that ∥B∥≤ and xB. Using this problem as an NP oracle, in \(\varTheta_{2}^{p}\) we can decide, given (A,≻) and dA, whether there exists a minimum-size upward covering set for A containing d as follows. The oracle is asked whether for each pair (x,), where xA and 1≤≤∥A∥, there exists an upward covering set for A of size bounded by that contains the alternative x. The number of queries is polynomial (more specifically in \(\mathcal{O}(\|A\|^{2})\)), and all queries can be asked in parallel. Having all the answers, determine the size k of a minimum-size upward covering set for A, and accept if the oracle answer to (d,k) was yes, otherwise reject.

To show that \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\) is in \(\varTheta_{2}^{p}\), let (A,≻) be a dominance graph and d a designated alternative in A. We now use as our oracle the set of all (x,), where xA is an alternative, and ≤∥A∥ a positive integer, such that there exists some upward covering set B for A with ∥B∥≤ and \(x \not\in B\). Clearly, this problem is also in NP, and the size k of a minimum-size upward covering set for A can again be determined by asking \(\mathcal{O}(\|A\|^{2})\) queries in parallel (if all oracle answers are no, it holds that k=∥A∥). Now, the \(\varTheta_{2}^{p}\) machine accepts its input ((A,≻),d) if the oracle answer for the pair (d,k) is no, and otherwise it rejects. □

Theorem 14

  1. 1.

    (Brandt and Fischer [7]) It is coNP-complete to decide whether a designated alternative is contained in all minimal upward covering sets for a given dominance graph. That is, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\)is coNP-complete.

     
  2. 2.

    It is coNP-complete to decide whether a given subset of the alternatives is a minimal upward covering set for a given dominance graph. That is, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\)is coNP-complete.

     
  3. 3.

    It is coNP-hard and in\(\varSigma_{2}^{p}\)to decide whether there is a unique minimal upward covering set for a given dominance graph. That is, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\)is coNP-hard and in\(\varSigma_{2}^{p}\).

     

Proof

It follows from Claim 6 that in Construction 3 the boolean formula φ is not satisfiable if and only if the entire set of alternatives A is a (unique) minimal upward covering set for A. Furthermore, if φ is satisfiable, there exists more than one minimal upward covering set for A and none of them contains e1 (provided that φ has more than one satisfying assignment, which can be ensured, if needed, by adding a dummy variable such that the satisfiability of the formula is not affected). This proves coNP-hardness for all three problems. \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\) and \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\) are also contained in coNP, as they can be decided in the positive by checking whether there does not exist an upward covering set that satisfies certain properties related to the problem at hand, so they both are coNP-complete. \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\) can be decided in the positive by checking whether there exists an upward covering set M such that all sets that are not strict supersets of M are not upward covering sets for the set of all alternatives. Thus, \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\) is in \(\varSigma_{2}^{p}\). □

The first statement of Theorem 14 was already shown by Brandt and Fischer [7]. However, their proof—which uses essentially the reduction from the proof of Theorem 2, except that they start from the coNP-complete problem Validity (which asks whether a given formula is valid, i.e., true under every assignment [41])—does not yield any of the other coNP-hardness results in Theorem 14.

Theorem 15

It is coNP-complete to decide whether a given subset of the alternatives is a minimum-size upward covering set for a given dominance graph. That is, \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\)is coNP-complete.

Proof

This problem is in coNP, since it can be decided in the positive by checking whether the given subset M of alternatives is an upward covering set for the set A of all alternatives (which is easy) and all sets of smaller size than M are not upward covering sets for A (which is a coNP predicate). Now, coNP-hardness follows directly from Claim 6, which shows that in Construction 3 the boolean formula φ is not satisfiable if and only if there is a unique minimal upward covering set for A and hence also a unique minimum-size upward covering set for A. □

Theorem 16

Deciding whether there exists a unique minimum-size upward covering set for a given dominance graph is hard for coNP and in \(\varTheta_{2}^{p}\). That is, \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\)is coNP-hard and in \(\varTheta_{2}^{p}\).

Proof

It is easy to see that coNP-hardness follows directly from the coNP-hardness of \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\) (see Theorem 14). Membership in \(\varTheta_{2}^{p}\) can be proven by using the same oracle as in the proof of the first part of Theorem 13. We ask for all pairs (x,), where xA and 1≤≤∥A∥, whether there is an upward covering set B for A such that ∥B∥≤ and xB. Having all the answers, determine the minimum size k of a minimum-size upward covering set for A. Accept if there are exactly k distinct alternatives x1,…,xk for which the answer for (xi,k), 1≤ik, was yes, otherwise reject. □

An important consequence of the proofs of Theorems 14 and 16 (and of Construction 3 that underpins these proofs) regards the hardness of the search problems \(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\).

Theorem 17

Assuming P≠NP, neither minimal upward covering sets nor minimum-size upward covering sets can be found in polynomial time. That is, neither\(\mbox {\textsc {MC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\)nor\(\mbox {\textsc {MSC}$_{\mathrm {u}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\)are polynomial-time computable unless P=NP.

Proof

Consider the problem of deciding whether there exists a nontrivial minimal/minimum-size upward covering set, i.e., one that does not contain all alternatives. By Construction 3 that is applied in proving Theorems 14 and 16, there exists a trivial minimal/minimum-size upward covering set for A (i.e., one containing all alternatives in A) if and only if this set is the only minimal/minimum-size upward covering set for A. Thus, the coNP-hardness proof for the problem of deciding whether there is a unique minimal/minimum-size upward covering set for A (see the proofs of Theorems 14 and 16) immediately implies that the problem of deciding whether there is a nontrivial minimal/minimum-size upward covering set for A is NP-hard. However, since the latter problem can easily be reduced to the search problem (because the search problem, when used as a function oracle, yields the set of all alternatives if and only if this set is the only minimal/minimum-size upward covering set for A), it follows that the search problem cannot be solved in polynomial time unless P=NP. □

5 Minimal and Minimum-Size Downward Covering Sets

Now we consider minimal and minimum-size downward covering sets.

5.1 Constructions

Again we first give the constructions that will be used in Sect. 5.2 to show complexity results about minimal/minimum-size downward covering sets. we again start by giving a proof sketch of a result due to Brandt and Fischer [7], since the following constructions and proofs are based on their construction and proof.

Theorem 18

(Brandt and Fischer [7])

Deciding whether a designated alternative is contained in some minimal downward covering set for a given dominance graph is NP-hard (i.e., \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)is NP-hard), even if a downward covering set is guaranteed to exist.

Proof Sketch

NP-hardness of \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) is again shown by a reduction from SAT. Given a boolean formula in conjunctive normal form, φ(v1,v2,…,vn)=c1c2∧…∧cr, over the set V={v1,v2,…,vn} of variables, construct a dominance graph (A,≻) as follows. The set of alternatives is
$$A= \bigl\{x_i,\overline{x}_i, x_i', \overline{x}_i',x_i'', \overline{x}_i'' \,\mid \:v_i \in V\bigr\} \cup \{y_j,z_j \,\mid \:c_j \mbox{ is a clause in } \varphi\} \cup \{d\}, $$
where the membership of alternative d in a minimal downward covering set is to be decided. The dominance relation ≻ is defined as follows:
  • For each i, 1≤in, there is a cycle \(x_{i} \succ\overline{x}_{i} \succ x_{i}' \succ\overline{x}_{i}' \succ x_{i}'' \succ\overline{x}_{i}'' \succ x_{i}\) with two nested three-cycles, \(x_{i} \succ x_{i}' \succ x_{i}'' \succ x_{i}\) and \(\overline{x}_{i} \succ \overline{x}_{i}' \succ\overline{x}_{i}'' \succ\overline{x}_{i}\);

  • if variable vi occurs in clause cj as a positive literal, then yjxi;

  • if variable vi occurs in clause cj as a negative literal, then \(y_{j} \succ\overline{x}_{i}\);

  • for each j, 1≤jr, we have dyj and zjd; and

  • for each i and j with 1≤i,jr and ij, we have ziyj.

Brandt and Fischer [7] showed that there is a minimal downward covering set containing d if and only if φ is satisfiable. An example of this reduction is shown in Fig. 6 for the boolean formula (v1∨¬v2v3)∧(¬v1∨¬v3). The set \(\{ x_{1},x_{1}',x_{1}'',x_{2},x_{2}',x_{2}'',\overline {x}_{3},\overline {x}_{3}',\overline {x}_{3}'',y_{1},y_{2},z_{1},z_{2},d\}\) is a minimal downward covering set for the dominance graph shown in Fig. 6. This set corresponds to the truth assignment that sets v1 and v2 to true and v3 to false, and it contains the designated alternative d.
Fig. 6

Dominance graph for Theorem 18, example for the formula (v1∨¬v2v3)∧(¬v1∨¬v3)

 □

Regarding their construction sketched above, Brandt and Fischer [7] showed that every minimal downward covering set for A must contain exactly three alternatives for every variable vi (either xi, \(x_{i}'\), and \(x_{i}''\), or \(\overline{x}_{i}\), \(\overline{x}_{i}'\), and \(\overline{x}_{i}''\)), and the undominated alternatives z1,…,zr. Thus, each minimal downward covering set for A consists of at least 3n+r alternatives and induces a truth assignment α for φ. The number of alternatives contained in any minimal downward covering set for A corresponding to an assignment α is 3n+r+k, where k is the number of clauses that are satisfied if α is an assignment not satisfying φ, and where k=r+1 if α is a satisfying assignment for φ. As a consequence, minimum-size downward covering sets for A correspond to those assignments for φ that satisfy the least possible number of clauses of φ.10

Next, we provide a different construction to transform a given boolean formula into a dominance graph. This construction will later be merged with the construction from the proof sketch of Theorem 18 so as to apply Lemma 1 to show \(\varTheta_{2}^{p}\)-hardness for downward covering set problems.

Construction 19

(for NP- and coNP-hardness of downward covering set problems)

Given a boolean formula in conjunctive normal form, φ(w1,w2,…,wk)=f1f2∧…∧f, over the setW={w1,w2,…,wk} of variables, we construct a dominance graph (A,≻). The set of alternatives is
$$A= A_1 \cup A_2 \cup\{\widehat {a}\,\mid \:a \in A_1 \cup A_2\} \cup \{b,c,d\} $$
with\(A_{1}=\{x_{i},x_{i}',x_{i}'',\overline{x}_{i},\overline{x}_{i}',\overline {x}_{i}'',z_{i},z_{i}',z_{i}'' \,\mid \:w_{i} \in W\}\)andA2={yj ∣ fjisaclauseinφ}, and the dominance relationis defined by:
  • For eachi, 1≤ik, there is, similarly to the construction in the proof of Theorem 18, a cycle\(x_{i} \succ\overline{x}_{i} \succ x_{i}' \succ\overline{x}_{i}' \succ x_{i}'' \succ\overline{x}_{i}'' \succ x_{i}\)with two nested three-cycles, \(x_{i} \succ x_{i}' \succ x_{i}'' \succ x_{i}\)and\(\overline{x}_{i} \succ \overline{x}_{i}' \succ\overline{x}_{i}'' \succ\overline{x}_{i}\), and additionally we have\(z_{i}' \succ z_{i} \succ x_{i}\), \(z_{i}'' \succ z_{i} \succ \overline{x_{i}}\), \(z_{i}' \succ x_{i}\), \(z_{i}'' \succ\overline{x}_{i}\), anddzi;

  • if variablewioccurs in clausefjas a positive literal, thenxiyj;

  • if variablewioccurs in clausefjas a negative literal, then\(\overline{x}_{i} \succ y_{j}\);

  • for eachaA1A2, we have\(b \succ \widehat {a}\), \(a \succ \widehat {a}\), and\(\widehat {a}\succ d\);

  • for eachj, 1≤j, we havedyj; and

  • cd.

An example of this construction is shown in Fig. 7 for the boolean formula (¬w1w2w3)∧(¬w2∨¬w3), which can be satisfied by setting for example each of w1, w2, and w3 to false. A minimal downward covering set corresponding to this assignment is \(M=\{b,c\} \cup\{\overline{x}_{i}, \overline{x}_{i}', \overline{x}_{i}'', z_{i}', z_{i}'' \,\mid \:1 \leq i \leq3\}\). Obviously, the undominated alternatives b, c, \(z_{i}'\), and \(z_{i}''\), 1≤i≤3, are contained in every minimal downward covering set for the dominance graph constructed. The alternative d, however, is not contained in any minimal downward covering set for A. This can be seen as follows. If d were contained in some minimal downward covering set M′ for A then none of the alternatives \(\widehat {a}\) with aA1A2 would be downward covered. Hence, all alternatives in A1A2 would necessarily be in M′, since they all dominate a different alternative in M′. But then M′ is no minimal downward covering set for A, since the minimal downward covering set M for A is a strict subset of M′.
Fig. 7

Dominance graph resulting from the formula (¬w1w2w3)∧(¬w2∨¬w3) according to Construction 19. An edge incident to a set of alternatives represents an edge incident to each alternative in the set. The dashed edge indicates that \(a\succ \widehat {a}\) for each aA1A2

We now show some properties of Construction 19 in general.

Claim 20

Minimal downward covering sets are guaranteed to exist for the dominance graph defined in Construction 19.

Proof

The set A of all alternatives is a downward covering set for itself. Hence, there always exists a minimal downward covering set for the dominance graph defined in Construction 19. □

Claim 21

Consider the dominance graph (A,≻) created by Construction 19. For each minimal downward covering set M for A, if M contains the alternative d then all other alternatives are contained in M as well (i.e., A=M).

Proof

If d is contained in some minimal downward covering set M for A, then \(\{a,\widehat {a}\} \subseteq M\) for every aA1A2. To see this, observe that for an arbitrary aA1A2 there is no a′∈A with \(a' \succ \widehat {a}\) and a′≻d or with a′≻a and \(a' \succ \widehat {a}\). Since the alternatives c and b are undominated, they are also in M, so M=A. □

Claim 22

Consider Construction 19. The boolean formula φ is satisfiable if and only if there is no minimal downward covering set for A that contains d.

Proof

For the direction from left to right, consider a satisfying assignment α:W→{0,1} for φ, and define the set It is not hard to verify that Bα is a minimal downward covering set for A. Thus, there exists a minimal downward covering set for A that does not contain d. If there were a minimal downward covering set M for A that contains d, Claim 21 would imply that M=A. However, since BαA=M, this contradicts minimality, so no minimal downward covering set for A can contain d.

For the direction from right to left, assume that no minimal downward covering set for A contains d. Since by Claim 20 minimal downward covering sets are guaranteed to exist for the dominance graph defined in Construction 19, there exists a minimal downward covering set B for A that does not contain d, so BA. It holds that {zi ∣ wi is a variable in φ}∩B=∅ and {yj ∣ fj is a clause in φ}∩B=∅, for otherwise a contradiction would follow by observing that there is no aA with ad and azi, 1≤ik, or with ad and ayj, 1≤j. Furthermore, we have \(x_{i} \not\in B\) or \(\overline{x}_{i} \not\in B\), for each variable wiW. By external stability, for each clause fj there must exist an alternative aB with ayj. By construction and since \(d \not\in B\), we must have either a=xi for some variable wi that occurs in fj as a positive literal, or \(a=\overline{x}_{i}\) for some variable wi that occurs in fj as a negative literal. Now define α:W→{0,1} such that α(wi)=1 if xiB, and α(wi)=0 otherwise. It is readily appreciated that α is a satisfying assignment for φ. □

Claim 23

Consider Construction 19. The boolean formula φ is not satisfiable if and only if there is a unique minimal downward covering set for A.

Proof

We again assume that if φ is satisfiable, it has at least two satisfying assignments. If φ is not satisfiable, there must be a minimal downward covering set for A that contains d by Claim 22, and by Claim 21 there must be a minimal downward covering set for A containing all alternatives. Hence, there is a unique minimal downward covering set for A. Conversely, if there is a unique minimal downward covering set for A, φ cannot be satisfiable, since otherwise there would be at least two distinct minimal downward covering sets for A, corresponding to the distinct truth assignments for φ, which would yield a contradiction. □

In the dominance graph created by Construction 19, the minimal downward covering sets for A coincide with the minimum-size downward covering sets for A. If φ is not satisfiable, there is only one minimal downward covering set for A, so this is also the only minimum-size downward covering set for A, and if φ is satisfiable, the minimal downward covering sets for A correspond to the satisfying assignments of φ. As we have seen in the proof of Claim 22, these minimal downward covering sets for A always consist of 5k+2 alternatives. Thus, they each are also minimum-size downward covering sets for A.

Merging the construction from the proof sketch of Theorem 18 with Construction 19, we again provide a reduction applying Lemma 1, this time to downward covering set problems.

Construction 24

(for applying Lemma 1 to downward covering set problems)

We again apply Wagner’s lemma with the NP-complete problemS=SAT and construct a dominance graph. Fix an arbitrarym≥1 and letφ1,φ2,…,φ2mbe 2mboolean formulas in conjunctive normal form such that the satisfiability ofφjimplies the satisfiability ofφj−1, for eachj∈{2,…,2m}. Without loss of generality, we assume that for each j, 1≤j≤2m, φjhas at least two satisfying assignments, ifφjis satisfiable.

We now define a polynomial-time computable functionf, which maps the given 2mboolean formulas to a dominance graph (A,≻) that has useful properties for our downward covering set problems. The set of alternatives is
$$A= \Biggl(\bigcup_{i=1}^{2m} A_i \Biggr) \cup \Biggl( \bigcup_{i=1}^m \{r_i,s_i,t_i \} \Biggr) \cup\bigl \{c^*,d^*\bigr\}, $$
and the dominance relationonAis defined bywhere we use the following notation:
  1. (1)

    For eachi, 1≤im, let (A2i−1,≻2i−1) be the dominance graph that results from the formulaφ2i−1according to Brandt and Fischer’s construction given in the proof sketch of Theorem 18. We again use the same names for the alternatives inA2i−1as in that proof sketch, except that we attach the subscript 2i−1.

     
  2. (2)

    For eachi, 1≤im, let (A2i,≻2i) be the dominance graph that results from the formulaφ2iaccording to Construction 19. We again use the same names for the alternatives inA2ias in that construction, except that we attach the subscript 2i.

     
  3. (3)

    For eachi, 1≤im, the dominance graphs (A2i−1,≻2i−1) and (A2i,≻2i) are connected by the alternativessi, ti, andri (which play a similar role as the alternativeszi, \(z_{i}'\), and\(z_{i}''\)for each variable in Construction 19). The resulting dominance graph is denoted by\((B_{i},\succ_{i}^{B})\).

     
  4. (4)

    Connect themdominance graphs\((B_{i},\succ_{i}^{B})\), 1≤im (again similarly as in Construction 19). The alternativecdominatesd, andddominates themalternativesri, 1≤im.

     
This construction is illustrated in Fig. 8. Clearly, (A,≻) is computable in polynomial time.
Fig. 8

Dominance graph from Construction 24

Claim 25

Consider Construction 24. For each i, 1≤i≤2m, let Mi be a minimal downward covering set for (Ai,≻i). Then each of the sets Mi must be contained in every minimal downward covering set for (A,≻).

Proof

For each i, 1≤i≤2m, the only alternative in Ai dominated from outside Ai is di. Since di is also dominated by the undominated alternative z1,iAi for odd i, and by the undominated alternative ciAi for even i, it is readily appreciated that internal and external stability with respect to elements of Ai only depends on the restriction of the dominance graph to Ai. □

Claim 26

Consider Construction 24. It holds that

Proof

For the direction from left to right in (5.3), assume that ∥{i ∣ φi∈SAT}∥ is odd. Thus, there is some j∈{1,…,m} such that φ1,φ2,…,φ2j−1 are each satisfiable and φ2j,φ2j+1,…,φ2m are each not. Define
$$M= \Biggl(\bigcup_{i=1}^{2m} M_i \Biggr) \cup \Biggl(\bigcup_{i=1}^m \{s_i,t_i \} \Biggr) \cup \bigl\{ r_j,c^*,d^* \bigr\}, $$
where for each i, 1≤i≤2m, Mi is some minimal downward covering set of the restriction of the dominance graph to Ai, satisfying that diMi if and only if
  1. 1.

    i is odd and φi is satisfiable, or

     
  2. 2.

    i is even and φi is not satisfiable.

     
Such sets Mi exist by the proof sketch of Theorem 18 and by Claim 22. In particular, φ2j−1 is satisfiable and φ2j is not, so {d2j−1,d2j}⊆M. There is no alternative that dominates d2j−1, d2j, and rj. Thus, rj must be in M. The other alternatives ri, 1≤im and ij, are downward covered by either si if \(d_{2i-1} \not\in M\), or ti if \(d_{2i} \not\in M\). Finally, d cannot be downward covered, because drj and no alternative dominates both d and rj. Internal and external stability with respect to the elements of Mi, as well as minimality of \(\bigcup_{i=1}^{2k} M_{i}\), follow from the proofs of Theorem 18 and Claim 22. All other elements of M are undominated and thus contained in every downward covering set. We conclude that M is a minimal downward covering set for A that contains d.

For the direction from right to left in (5.3), assume that there exists a minimal downward covering set M for A with dM. By internal stability, there must exist some j, 1≤jk, such that rjM. Thus, d2j−1 and d2j must be in M, too. It then follows from the proof sketch of Theorem 18 and Claim 22 that φ2j−1 is satisfiable and φ2j is not. Hence, ∥{i ∣ φi∈SAT}∥ is odd. □

By the remark made after Theorem 18, Construction 24 cannot be used straightforwardly to obtain complexity results for minimum-size downward covering sets.

5.2 Proofs

Now we prove the remaining parts of Theorem 1 concerning minimal and minimum-size downward covering sets by applying the constructions and the properties of the resulting dominance graphs presented in Sect. 5.1.

Theorem 27

It is NP-complete to decide, given a dominance graph (A,≻) and a positive integer k, whether there exists a minimal/minimum-size downward covering set forAof size at most k. That is, \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\)and\(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\)are both NP-complete.

Proof

Membership in NP is obvious, since we can nondeterministically guess a subset MA of the alternatives with ∥M∥≤k and can then check in polynomial time whether M is a downward covering set for A. NP-hardness of \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Size}}\) follows from Construction 19, the proof of Claim 22, and the comments made after Claim 23: If φ is a given formula with n variables, then there exists a minimal/minimum-size downward covering set of size 5n+2 if and only if φ is satisfiable. □

Theorem 28

\(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\), \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\), and\(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\)are coNP-hard and in \(\varTheta_{2}^{p}\).

Proof

It follows from Claim 23 that in Construction 19 the boolean formula φ is not satisfiable if and only if the entire set A of all alternatives is the unique minimum-size downward covering set for itself. Moreover, assuming that φ has at least two satisfying assignments, if φ is satisfiable, there are at least two distinct minimum-size downward covering sets for A. This shows that each of \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\), \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\), and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\) is coNP-hard. For all three problems, membership in \(\varTheta_{2}^{p}\) is shown similarly to the proofs of the corresponding minimum-size upward covering set problems. However, since downward covering sets may fail to exist, the proofs must be slightly adapted. For \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\), the machine rejects the input if the size k of a minimum-size downward covering set cannot be computed (simply because there doesn’t exist any such set). For \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\), if all oracle answers are no, it must be checked whether the set of all alternatives is a downward covering set for itself. If so, the machine accepts the input, otherwise it rejects. □

Theorem 29

It is coNP-complete to decide whether a given subset is a minimum-size downward covering set for a given dominance graph. That is, \({\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}}\)is coNP-complete.

Proof

This problem is in coNP, since its complement (i.e., the problem of deciding whether a given subset of the set A of alternatives is not a minimum-size downward covering set for A) can be decided in nondeterministic polynomial time. Hardness for coNP follows directly from Claim 23, which shows that in Construction 19 the boolean formula φ is not satisfiable if and only if there is a unique minimal downward covering set for A and hence also a unique minimum-size downward covering set for A. □

Theorem 30

Deciding whether a designated alternative is contained in some minimal downward covering set for a given dominance graph is hard for\(\varTheta_{2}^{p}\)and in\(\varSigma_{2}^{p}\). That is, \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\)is hard for\(\varTheta_{2}^{p}\)and in \(\varSigma_{2}^{p}\).

Proof

Membership in \(\varSigma_{2}^{p}\) can be shown analogously to the proof of Theorem 12, and \(\varTheta_{2}^{p}\)-hardness follows directly from Claim 26, which applies Wagner’s lemma to downward covering sets. Specifically, this claim shows that in Construction 24 the alternative d is contained in some minimal downward covering set for A if and only if the number of underlying boolean formulas is odd. □

Theorem 31

  1. 1.

    (Brandt and Fischer [7]) It is coNP-complete to decide whether a designated alternative is contained in all minimal downward covering sets for a given dominance graph. That is, \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\)is coNP-complete.

     
  2. 2.

    It is coNP-complete to decide whether a given subset of the alternatives is a minimal downward covering set for a given dominance graph. That is, \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\)is coNP-complete.

     
  3. 3.

    It is coNP-hard and in\(\varSigma_{2}^{p}\)to decide whether there is a unique minimal downward covering set for a given dominance graph. That is, \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\)is coNP-hard and in\(\varSigma_{2}^{p}\).

     

Proof

It follows from Claim 23 that in Construction 19 the boolean formula φ is not satisfiable if and only if the entire set of alternatives A is a unique minimal downward covering set for A. Furthermore, if φ is satisfiable, there exists more than one minimal downward covering set for A and none of them contains d (provided that φ has more than one satisfying assignment, which can be ensured, if needed, by adding a dummy variable such that the satisfiability of the formula is not affected). This proves coNP-hardness for all three problems. \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Member}}\hbox {-}\allowbreak \mbox {\textsc {All}}\) and \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Test}}\) are also contained in coNP, because they can be decided in the positive by checking whether there does not exist a downward covering set that satisfies certain properties related to the problem at hand. Thus, they are both coNP-complete. \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\) can be decided in the positive by checking whether there exists a downward covering set M such that all sets that are not strict supersets of M are not downward covering sets for the set of all alternatives. This shows that \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Unique}}\) is in \(\varSigma_{2}^{p}\). □

The first statement of Theorem 31 was already shown by Brandt and Fischer [7]. However, their proof—which uses essentially the reduction from the proof of Theorem 18, except that they start from the coNP-complete problem Validity—does not yield any of the other coNP-hardness results in Theorem 31.

An important consequence of the proofs of Theorems 28 and 31 regards the hardness of the search problems \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) and \(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\). (Note that the hardness of \(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\) also follows from a result by Brandt and Fischer [7, Theorem 9], see the discussion in Sect. 3.)

Theorem 32

Assuming P≠NP, neither minimal downward covering sets nor minimum-size downward covering sets can be found in polynomial time (i.e., neither\(\mbox {\textsc {MC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\)nor\(\mbox {\textsc {MSC}$_{\mathrm {d}}$}\hbox {-}\allowbreak \mbox {\textsc {Find}}\)are polynomial-time computable unless P=NP), even when the existence of a downward covering set is guaranteed.

Proof

Consider the problem of deciding whether there exists a nontrivial minimal/minimum-size downward covering set, i.e., one that does not contain all alternatives. By Construction 19 that is applied in proving Theorems 28 and 31, there exists a trivial minimal/minimum-size downward covering set for A (i.e., one containing all alternatives in A) if and only if this set is the only minimal/minimum-size downward covering set for A. Thus, the coNP-hardness proof for the problem of deciding whether there is a unique minimal/minimum-size downward covering set for A (see the proofs of Theorems 28 and 31) immediately implies that the problem of deciding whether there is a nontrivial minimal/minimum-size downward covering set for A is NP-hard. However, since the latter problem can easily be reduced to the search problem (because the search problem, when used as a function oracle, yields the set of all alternatives if and only if this set is the only minimal/minimum-size downward covering set for A), it follows that the search problem cannot be solved in polynomial time unless P=NP. □

6 Conclusions and Open Questions

In this paper we have systematically studied the complexity of various problems related to inclusion-minimal and minimum-size unidirectional (i.e., either upward or downward) covering sets. We have established hardness or completeness results for either of NP, coNP, and \(\varTheta_{2}^{p}\) (see Tables 1 and 2 in Sect. 3). An important consequence is that if P≠NP then neither minimal upward nor minimal downward covering sets (even when guaranteed to exist) can be computed in polynomial time. This sharply contrasts with Brandt and Fischer’s result that minimal bidirectional covering sets in fact are polynomial-time computable [7].

Tables 1 and 2 also list the best upper bounds we could establish for these problems. In some cases, these upper bounds do not coincide with the lower bounds established, for example, when \(\varTheta_{2}^{p}\)-hardness but only membership in \(\varSigma_{2}^{p}\) could be proven. As an interesting task for future research, we propose to close these complexity gaps. As suggested by an anonymous reviewer, a good candidate problem for finding a reduction to prove \(\varSigma_{2}^{p}\)-completeness for problems related to minimal unidirectional covering sets is the problem of deciding whether a positive literal belongs to a minimal model of a propositional formula (see [18]).

Footnotes
1

Such payoff vectors are called imputations; see, e.g., [12, 40] for the game-theoretic notions not defined here.

 
2

In general, ≻ need not be transitive or complete. For alternatives x and y, xy (equivalently, (x,y)∈ ≻) is interpreted as x being strictly preferred to y (and we say “x dominates y”), e.g., due to a strict majority of voters preferring x to y (recall Fig. 1 for an example).

 
3

Consider the set A={a,b,c} of three alternatives with the dominance relation defined by abc. Note that A is not a downward covering set for itself, since it violates internal stability (UCd(A)={a,b}≠A, due to c being downward covered by b in A); both {a,b} and {b,c} violate internal stability as well (e.g., UCd({a,b})={a}≠{a,b}); and external stability is violated by {a,c} (due to b∈UCd({a,c}∪{b})=UCd(A)={a,b}), each singleton (c∈UCd({a}∪{c})={a,c} shows this for {a}, a∈UCd({b}∪{a})={a} works for {b}, and b∈UCd({c}∪{b})={b} works for {c}), and the empty set (due to, e.g., a∈UCd(∅∪{a})={a}). Thus A has no downward covering set at all.

 
4

Consider, for example, the set A={a,b,c,d,e} of five alternatives with the dominance relation defined by abcda and be. It is easy to see that both {a,c,e} and {b,d} are minimal upward covering sets for A, but only {b,d} is an upward covering set of minimum size for A. That is, {a,c,e} is a minimal, but not minimum-size upward covering set for A.

 
5

This type of reduction was introduced by Ladner et al. [32]. Informally stated, a disjunctive truth-table reduction between two decision problems X and Y computes, given an instance x, in polynomial time k queries y1,y2,…,yk such that xX if and only if yiY for at least one i, 1≤ik. This reduction can be adapted straightforwardly to function problems F and G: Fdisjunctively truth-table reduces toG if, given an instance x, in polynomial time we can compute k queries y1,y2,…,yk such that F(x) can be computed from G(yi) for at least one i, 1≤ik.

 
6

The argument is analogous to that used in the construction of Brandt and Fischer [7] in their proof of Theorem 2. However, in contrast with their construction, which implies that either\(\{x_{i}, x'_{i}\}\)or\(\{\overline{x}_{i}, \overline{x}'_{i}\}\), 1≤in, but not both, must be contained in any minimal upward covering set for A (see Fig. 2), our construction also allows for both \(\{u_{i}, u'_{i}\}\) and \(\{\overline{u}_{i}, \overline{u}'_{i}\}\) being contained in some minimal upward covering set for A. Informally stated, the reason is that, unlike the four-cycles in Fig. 2, our four-cycles \(u_{i} \succ\overline{u}_{i} \succ u'_{i} \succ\overline{u}'_{i} \succ u_{i}\) also have incoming edges.

 
7

For example, recall Wagner’s \(\varTheta_{2}^{p}\)-completeness result for testing whether the size of a maximum clique in a given graph is an odd number [51]. One key ingredient in his proof is to define an associative operation on graphs, ⋈, such that for any two graphs G and H, the size of a maximum clique in GH equals the sum of the sizes of a maximum clique in G and one in H. This operation is quite simple: Just connect every vertex of G with every vertex of H. In contrast, since minimality for minimal upward covering sets is defined in terms of set inclusion, it is not at all obvious how to define a similarly simple operation on dominance graphs such that the minimal upward covering sets in the given graphs are related to the minimal upward covering sets in the connected graph in a similarly useful way.

 
8

Our argument about \((B_{i}, \succ_{i}^{B})\) can be used to show, in effect, DP-hardness of upward covering set problems, where DP is the class of differences of any two NP sets [42]. Note that DP is the second level of the boolean hierarchy over NP (see Cai et al. [10, 11]), and it holds that \(\mathrm {NP}\cup \mathrm {coNP}\subseteq \mathrm {DP}\subseteq \varTheta_{2}^{p}\). Wagner [51] proved appropriate analogs of Lemma 1 for each level of the boolean hierarchy. In particular, the analogous criterion for DP-hardness is obtained by using the wording of Lemma 1 except with the value of m=1 being fixed.

 
9

This implies that d1 is not upward covered by either \(u_{1,2}'\) or \(\overline{u}_{1,2}'\), since d3 dominates them both but not d1.

 
10

This is different from the case of minimum-size upward covering sets for the dominance graph constructed in the proof sketch of Theorem 2. The construction in the proof sketch of Theorem 18 cannot be used to obtain complexity results for minimum-size downward covering sets in the same way as the construction in the proof sketch of Theorem 2 was used to obtain complexity results for minimum-size upward covering sets.

 

Acknowledgements

This work was supported in part by DFG grants BR-2312/6-1, RO-1202/12-1 (within the European Science Foundation’s EUROCORES program LogICCC), BR 2312/3-2, RO-1202/11-1, and RO-1202/15-1, and by the Alexander von Humboldt Foundation’s TransCoop program. This work was done in part while the fifth author was visiting the University of Rochester.

Copyright information

© Springer Science+Business Media New York 2012

Authors and Affiliations

  • Dorothea Baumeister
    • 1
  • Felix Brandt
    • 2
  • Felix Fischer
    • 3
  • Jan Hoffmann
    • 4
  • Jörg Rothe
    • 1
  1. 1.Institut für InformatikHeinrich-Heine-Universität DüsseldorfDüsseldorfGermany
  2. 2.Institut für InformatikTechnische Universität MünchenGarchingGermany
  3. 3.Statistical LaboratoryUniversity of CambridgeCambridgeUK
  4. 4.Department of Computer ScienceYale UniversityNew HavenUSA

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