Ranking Specific Sets of Objects

Ranking sets of objects based on an order between the single elements has been thoroughly studied in the literature. In particular, it has been shown that it is in general impossible to find a total ranking – jointly satisfying properties as dominance and independence – on the whole power set of objects. However, in many applications certain elements from the entire power set might not be required and can be neglected in the ranking process. For instance, certain sets might be ruled out due to hard constraints or are not satisfying some background theory. In this paper, we treat the computational problem whether an order on a given subset of the power set of elements satisfying different variants of dominance and independence can be found, given a ranking on the elements. We show that this problem is tractable for partial rankings and NP-complete for total rankings.


Introduction
The problem of lifting rankings on objects to ranking on sets has been studied from many different view points -see [2] for an excellent survey. Several properties (also called axioms) have been proposed in order to indicate whether the lifted ranking reflects the given order on the elements. Two important axioms are dominance and independence. Roughly speaking, dominance ensures that adding an element which is better (worse) than all elements in a set, Jan Maly jmaly@dbai.tuwien.ac.at 1 Institute of Information Systems, TU Wien, Wien, Austria makes the augmented set better (worse) than the original one. Independence, on the other hand, states that adding an element a to sets A and B where A is already known to be preferred over B, must not make B [ fag be preferred over A[fag (or, in the strict variant, A[fag should remain preferred over B [fag). These axioms were first considered together in the context of decision making under complete uncertainty [9]. There, sets represent the (mutually exclusive) possible outcomes of an action and one tries to rank these sets based on a preference ranking on the outcomes. It is assumed that the probability of each outcome is unknown, i. e., it is only known whether an event is a possible outcome or not. This is a very reductive model. Still, "it does succeed in modelling some empirically interesting situations" [5, p. 2]. Especially, "when the number of possible states of the world is large, an agent of bounded rationality may be incapable of undertaking (or unwilling to undertake) the complex calculations which consideration of the entire rows in the outcome matrix will involve." [15, p. 2]. Such situations often occur for autonomous agents, for example self driving cars, where "the temporal evolution of situations cannot be predicted without uncertainty because other road users behave stochastically and their goals and plans cannot be measured" [6, p. 1]. Moreover, dominance and independence are also sensible axioms in other contexts, for example bundles of objects of unknown size. Finally, to mention a very different application, Packard [14] used independence and a version of dominance to define plausibility rankings on theories.
However, it is well known that constructing a ranking on the whole power set of objects which jointly satisfies dominance and (strict) independence is, in general, not possible.

Example 1
Consider the problem of assigning tasks to agents. Let X = ft 1 ; :::; t n g be a collection of tasks. Further-more, assume we know for every agent what tasks they prefer to perform. If there are more tasks than agents, some agents have to perform several tasks, therefore it would be useful to know the preferences over sets of tasks. However, asking for these preferences directly is infeasible even for a reasonable small number of tasks. Therefore, we would like to lift the preferences over tasks to preferences over sets. Furthermore, it seems reasonable that the order on the sets should satisfy dominance and (strict) independence. Unfortunately, for strict independence, this is impossible even for n = 3. Assume t 1 < t 2 < t 3 . Then, ft 1 g ft 1 ; t 2 g is implied by dominance, therefore ft 1 ; t 3 g ft 1 ; t 2 ; t 3 g must hold by strict independence. On the other hand, ft 2 ; t 3 g ft 3 g is also implied by dominance, therefore, ft 1 ; t 2 ; t 3 g ft 1 ; t 3 g by strict independence. We thus end up with ft 1 ; t 2 ; t 3 g ft 1 ; t 3 g ft 1 ; t 2 ; t 3 g, hence is not an order.
Because of this, other (weaker) axiomatizations were proposed (see for example [7], or more recently, [4] and [10] among many others). However, in many applications one does not need to order the entire power set (for example, some tasks cannot be performed in parallel). In these cases, it may be possible to construct rankings that jointly satisfy dominance and (strict) independence.
Example 2 Let X be as above. Now assume ft 1 ; t 2 ; t 3 g is not a possible combination of tasks, for example, because fulfilling all three tasks at once is not feasible. Then, for example ft 1 g ft 1 ; t 2 g ft 2 g ft 1 ; t 3 g ft 2 ; t 3 g ft 3 g is a total order that satisfies dominance and strict independence (respecting the underlying linear order t 1 < t 2 < t 3 ).
In this paper, we investigate exactly this situation, i. e., lifting rankings to specific sets of elements. In the literature, this scenario seems to be rather neglected, so far. The only exception we are aware of deals with subsets of a fixed cardinality [3]. In particular, we are interested in the complexity of computing, if possible, rankings on arbitrary subsets of the power set that satisfy dominance and (strict) independence. To do so, we first give a new definition for dominance which appears more suitable in such a setting (for more details, see Section 3). Then, we consider the following problem: Given a ranking on elements, and a set S of sets of elements, does there exist a strict (partial) order on S that satisfies D and I (where D is either standard dominance or our notion of dominance and I is independence or strict independence)? We show that the problem is either trivial or easy to solve for the case of partial orders. Our main result is NP-completeness for the case when total orders are required.
The remainder of the paper is organized as follows. In the next section, we recall some basic concepts. In Section 3 we discuss why standard dominance can be seen as too weak in our setting and propose an alternative definition. Section 4 contains our main results. We conclude the paper in Section 5 with a summary and pointers to future work. This paper is an extended version of [11].

Background
The formal framework we want to consider in the following consists of a finite 1 nonempty set X, equipped with a linear order < and a subset X Â P.X/nf¿g of the power set of X not containing the empty set. We want to find a binary relation on X that satisfies some niceness conditions. We will consider several kinds of relations. We recall the relevant definitions.
Definition 1 A binary relation is called a strict partial order, if it is irreflexive and transitive. A strict or linear order is a total strict partial order. A binary relation is called a preorder, if it is reflexive and transitive. A (weak) order is a total preorder. If is a weak or a preorder on a set X, for all x; y 2 X, the corresponding strict order is defined by x y if x y and y 6 x hold.
Additionally, we need the following notions: Definition 2 For a pre-or weak order , we write x y if x y and y x hold. Let A 2 X be a set of elements of X. Then we write max.A/ for the maximal element of A with respect to < and min.A/ for the minimal element of A with respect to <. Furthermore, we say a relation R on a set X extends a relation S on X if xSy implies xRy for all x; y 2 X . Finally, we say a relation R on X is the transitive closure of a relation S on X if the existence of a sequence x 1 Sx 2 S :::S x k implies x 1 Rx k for all x 1 ; x k 2 X and R is the smallest relation with this property. We write trcl.S/ for the transitive closure of S.
Many different axioms a good order should satisfy are discussed in the literature (an overview over the relevant interpretations and the corresponding axioms can be found in the survey [2]). The following axioms "have very plausible intuitive interpretations" [2, p. 11] for decision making under complete uncertainty and belong to the most extensively studied ones. (We added conditions of the form X 2 X that are not necessary if X = P.X/nf¿g holds.)
Every reasonable order should satisfy the extension rule. If we assume X = P.X/nf¿g, the extension rule is implied by dominance [2]. Therefore, a natural task is to find a total order on P.X/nf¿g that satisfies dominance together with (some version of) independence. However, in their seminal paper [9], Kannai and Peleg have shown that this is impossible for regular independence and dominance if jXj 6 and X = P.X/nf¿g hold. Barberà and Pattanaik [1] showed that for strict independence and dominance this is impossible even for jXj 3 and X = P.X/nf¿g (see Example 1 for a proof of the statement).
If we abandon the condition X = P.X/nf¿g, the situation is not as clear. As we have seen in Example 2 there are sets X Â P.X/nf¿g with jXj 3 such that there is an order on X satisfying strict independence and dominance.

A Stronger Form of Dominance
Many results regularly used in the setting of X = P.X/nf¿g are not true in the more general case. For example, in contrast to the result stated above, the extension rule is not implied by dominance as we have seen in Example 3. Furthermore, it could be argued that f1,3g f1,4g should hold in that example which would be implied by dominance and independence if f3,4g 2 X would hold, because f3,4g f4g holds by dominance and so f1; 3; 4g f1,4g by independence. Hence, f1,3g f1; 3; 4g f1,4g implies f1,3g f1,4g by transitivity.
Furthermore, for the set X from Example 3, dominance does not even imply f1g f1; 2; 3g if f1,2g is not in the family. Therefore, it is reasonable to ask for a stronger version of dominance that behaves nicely in the general case. We observe that x < y for all y 2 A implies max.A [ fxg/ = max.A/ and min.A [ fxg/ < min.A/; whereas y < x for all y 2 A implies max.A/ < max.A [ fxg/ and min.A [ fxg/ = min.A/. We claim that every dominance-like axiom should satisfy this property. Therefore, we can use this property to define a "maximal" version of dominance which can be seen as a special case of Pareto dominance [12].

Axiom 5 (Maximal Dominance) For all
This axiom trivially implies the extension rule and of course dominance. Looking again at the family introduced in Example 3 maximal dominance implies all preferences implied by either dominance or by the extension rule and additionally f1,3g f1,4g. Furthermore, if X is sufficiently large or even X = P.X/nf¿g, dominance and independence imply maximal dominance.

Proposition 1 Let X = P.X/nf¿g. Then every transitive relation that satisfies dominance and independence also satisfies maximal dominance and independence.
Proof Let be a transitive relation that satisfies dominance and independence. We show that satisfies maximal dominance using the following observation due to Kannai and Peleg in [9]: We can assume w.l.o.g. that A has more than two elements. We enumerate A by A = fa 1 ; a 2 ; :::; a k g such that a i < a j holds for all i < j . Using transitivity and dominance, it is easy to see that fa 1 g fa 1 ; a 2 ; :::; a k−1 g holds. This implies, by independence, fa 1 ; a k g A. Analogously, we get fa 2 ; a 2 ; :::; a k g fa k g and A fa 1 ; a k g and therefore A fa 1 ; a k g = fmin.A/; max.A/g.U sing this observation we can prove that max.A/ = max.B/ and min.A/ < min.B/ implies A B by the following argument: The other case is proven analogously, hence satisfies maximal dominance.
It would be possible to define several other versions of dominance of intermediate strength. We will only consider dominance and maximal dominance. As we will see our results justify this approach; in particular, since both versions yield equal complexity results.

Main Results
We studied 8 problems in total, as defined below. 2 Our results are summarized in Table 1.

Problem 1 (The Partial (Maximal) Dominance Strict Independence problem)
Given a linearly ordered set X and a set X Â P.X/nf¿g, decide if there is a partial order on X satisfying (maximal) dominance and strict independence.

Problem 2 (The Partial (Maximal) Dominance Independence problem)
Given a linearly ordered set X and a set X Â P.X/nf¿g, decide if there is a preorder on X satisfying (maximal) dominance and independence. Problem 3 (The (Maximal) Dominance Strict Independence problem) Given a linearly ordered set X and a set X Â P.X/nf¿g, decide if there is a strict total order on X satisfying (maximal) dominance and strict independence.

Problem 4 (The (Maximal) Dominance Independence problem)
Given a linearly ordered set X and a set X Â P.X/nf¿g, decide if there is a total order on X satisfying (maximal) dominance and independence.

Partial Orders
First, we consider the Partial (Maximal) Dominance Independence problem. We can define a preorder that satisfies independence and maximal dominance (and therefore also dominance) on all X . Definition 3 Given a set X, a linear order < on X and a family X Â P.X/n¿, we define a relation m as A m B iff max.A/ Ä max.B/ and min.A/ Ä min.B/.
Observe that it is obviously possible, given X, < and X , to construct m in polynomial time. Example 4 Consider once again the family from Example 3. m consists of the following preferences on that family: f1,3g m f1; 2; 3g m f1; 3; 4g m f1,4g m f4g; f1,3g m f1; 2; 3g m f2,3g m f3g m f4g: Next, we consider the Partial Dominance Strict Independence problem. As we have seen in Example 1 and 2, only some sets X allow such an order. In order to decide if a set admits a partial order we build a minimal transitive relation satisfying dominance and strict independence.
K First, we build a minimal transitive relation satisfying dominance. It is worth noting that a very similar relation can be defined for maximal dominance. With this relation, all results in this section can be proven for maximal dominance the same way. Definition 4 Given a set X, a linear order < on X and a family X Â P.X/n¿, we define a relation d on X in the following way: If A; A [ fxg 2 X , then This relation has the following useful property.

Proposition 2
For every linearly ordered set X and every family X Â P.X/n¿, t d is a partial order and a partial order on X satisfies dominance if and only if it extends t d . Proof By definition, a relation satisfies dominance if and only if it extends d and a transitive relation extending d also extends t d by the minimality of trcl.
We want to extend this relation to a minimal relation for strict independence and dominance.
Definition 5 Given a set X, a linear order < on X and a family X Â P.X/n¿, we build a relation 1 on X by induction. First, we set t 0 W= t d . Now let t n be defined. For n+1 we select sets A; B; Anfxg; Bnfxg 2 X with x 2 X, Anfxg t n Bnfxg but not A n t B and set C n+1 D if C t n D or C = A and D = B holds. Then, we set t n+1 W= trcl. n+1 /. In the end, we set 1 = S n t n . In order to prove that this is actually a minimal order for dominance and strict independence, we have to introduce another concept we call links. Definition 6 A 1 -link from A to B in X is a sequence A =W C 0 ; C 1 ; :::; C n W= B with C i 2 X for all i Ä n such that, for all i < n, either C i d C i +1 holds or there is a link between C i nfxg and C i +1 nfxg for some x 2 X.
We show that 1 -links indeed characterize 1 . A; B 2 X , A 1 B implies that there is a   1 -link from A to B and if there is a 1 -link from A to B then A B holds for every transitive relation that satisfies dominance and strict independence.

Lemma 1 For
In order to prove this result, we need another definition.

Definition 7
For every pair A 1 B, there is a minimal k such that A t k B holds. We call this the 1 -rank of the pair.

Now we can prove Lemma 1:
Proof Assume A 1 B. We prove that a 1 -link exists by induction on the 1 -rank of A; B. If A t d B, then there is sequence A = C 1 ; C 2 ; :::; C n = B such that C i d C i +1 holds for all i < n, hence there is a 1 -link from A to B. Now assume A; B has 1 -rank k and for every pair with 1 -rank k − 1 or less there is a 1 -link from C to D. There is a sequence A = C 0 k C 1 :::C n−1 k C n = B. For every i < n either C i d C i +1 or C i nfyg t k−1 C i +1 nfyg holds, which implies by induction that there is a 1 -link from C i nfyg to C i +1 nfyg. Hence there is a 1 -link from A to B. Now, let be a transitive relation that satisfies dominance and strict independence and assume there is a 1link A = C 1 ; C 2 ; :::; C n = B from A to B. We prove A B by induction on the rank of the 1 -link. First, assume rank.C 1 ; C 2 ; :::; C n / = 1, then C i d C i +1 holds for all i < n, hence A B holds by dominance and transitivity. Now assume rank.C 1 ; C 2 ; :::; C n / = k and for all 1 -links with rank.C 1 ; C 2 ; :::; C n / < k we know C 1 C n . By induction, for every i < n either C i d C i +1 or C i nfxg C i +1 nfxg holds. This implies that C i C i +1 holds for all i < n, because satisfies dominance and strict independence. Therefore A B by transitivity.
Using this lemma, we can show now that 1 is indeed a minimal relation for dominance and strict independence.

Proof
1 satisfies dominance as it extends t d . By construction it also satisfies strict independence and transitivity: A 1 1 A 2 1 ::: 1 A k implies A 1 t n A 2 t n ::: t n A k for some n 2 N but then A 1 t n A k holds by the transitivity of t n and therefore A 1 1 A k . Now assume A 1 B and hence A t n B for some n and A [ fxg6 t n B [ fxg 2 X for some On the other hand, if 1 is not irreflexive no strict partial order can extend it. But every strict partial order on X satisfying dominance and strict independence must be an extension of 1 . Assume otherwise there is a strict partial order on X satisfying dominance and strict independence that does not extend 1 , i. e., there are sets A; B 2 X such that A 1 B holds but not A B. By Lemma 1 there is a 1 -link from A to B. This implies, by Lemma 1, A B because is transitive and satisfies dominance and strict independence. Contradiction. Therefore no partial order on X can satisfy dominance and strict independence, if 1 is irreflexive.
Using this result, we can define a polynomial time algorithm for the Partial Dominance Strict Independence Problem.

Corollary 1 The Partial Dominance Strict Independence problem is in P .
Proof Computing 1 can obviously be done in polynomial time because the construction always stops after at most jn nj = n 2 steps. Then checking if 1 is irreflexive only requires checking if A 1 A holds for some A.
Finally, links give us an easy characterization of sets X for which 1 is irreflexive.

Proof
1 is transitive and satisfies dominance and strict independence, hence Lemma 1 tells us, that A 1 A if and only if there is a 1 link from A to A.

Total Orders
We show that it is, in general, not possible to construct a (strict) total order satisfying both (maximal) dominance and (strict) independence deterministically in polynomial time 3 . We do this by a reduction from betweenness.

Problem 5 (Betweenness)
Given a set V = fv 1 ; v 2 ; :::; v n g and a set of triples R Â V 3 , does there exist a strict total order on V such that a < b < c or a > b > c holds for all .a; b; c/ 2 R.
Betweenness is known to be NP-hard [13]. We use this result to show NP-hardness for all four versions of the (Maximal) Dominance (Strict) Independence problem. The idea is, roughly, to represent the elements of V by sets Fig. 4 Sketch of the sets A; B; Anfkg; Bnfk + 1g; Anf1; k; k + 4g and Bnf1; k + 1; k + 4g which are not directly comparable via the axioms of dominance or independence. Hence, in order to find a total order, we need to guess how these sets are ordered. Starting from this guess we need to "maximize" this initial order in such a way that for each triple .a; b; c/ both a < b > c and a > b < c would lead to a circle in every order satisfying dominance and independence. However, this requires a number of carefully chosen additional sets as we will detail below.

Theorem 4.3 The Maximal Dominance Strict Independence problem, the Dominance Strict Independence problem, the Maximal Dominance Independence problem and the Dominance Independence problem are NP-complete.
It is clear that all four problems are in NP. We can guess a binary relation and then check if it has all properties we want. It is well known that checking for transitivity and (ir-) reflexivity can be done in polynomial time. Checking (maximal) dominance only requires an easy check for every pair of sets and (strict) independence an equally easy check for every quadruple of sets. It is clear that this can be done in polynomial time. In what follows, we split the proof of the NP-hardness in four parts, one for each problem.

The Maximal Dominance Strict Independence problem
Proof Let .V; R/ be an instance of betweenness with V = fv 1 ; v 2 ; :::; v n g. We construct an instance .X; <; X / of the Maximal Dominance Strict Independence problem. We set X = f1; 2; :::; N g equipped with the usual linear order, for N = 8n 3 + 2n + 2. Then, we construct the family X stepwise. The family contains for every v i 2 V a set V i of the following form (see Figure 3): V i W= f1; N g [ fi + 1; i + 2; :::; N − ig: Furthermore, for every triple from R we want to enforce A B C or A B C by adding two families of sets as shown in Figure 1 and Figure 2 with q; x; y; z 2 X. The solid arrows represent preferences that are forced through maximal dominance and strict independence. The family in Figure 1 makes sure that every total strict order satisfying independence that contains A B must also contain B C . Similarly, the family in Figure 2 makes sure that A B leads to B C .
We repeat this with a new triple .v 0 i ; v 0 j ; v 0 m / 2 R until we treated all triples in R. Observe that there are at most n 3 triples, thus, for every triple, the values k; :::; k + 7 lie between n + 1 and N − n, hence are element of every V i . In total, we add 24 sets per triple. Therefore, X contains n + 24n 3 sets.
It is easy to see, that, by construction, for every strict total order on X satisfying maximal dominance and strict independence, we have Anfkg Bnfk + 1g; Cnfk + 1g Bnfkg; Bnfk + 3g Anfk + 2g; Bnfk + 2g C nfk + 3g: Now assume there is a strict total order on X satisfying maximal dominance and strict independence. We claim that the relation defined by v i < v j iff V i V j is a positive witness for .V; R/. By definition this is a strict total order. So assume there is a triple .a; b; c/ such that a > b < c or a < b > c holds. We treat the first case in detail: a > b < c implies A B C . This implies by the strictness of and strict dominance Anfkg Bnfkg and Bnfk + 1g C nfk + 1g. However, then Anfkg Bnfkg C nfk + 1g Bnfk + 1g Anfkg contradicts the assumption that is transitive and irreflexive. Similarly, the second case leads to a contradiction. Now assume that there is a strict total order on V satisfying the restrictions from R. We use this to construct an order on X . We set V i V j iff v i < v j holds. Furthermore, we set A B for all A; B 2 X if it is implied by dominance. Then, we apply strict independence twice and once "reverse" strict independence 5 , i.e., A B implies Anfxg Bnfxg for A; B; Anfxg; Bnfxg 2 X . We claim that all possible instances of strict independence are decided already by this order. If A = V i for i Ä n, then there is no set A [ fxg in X . If A = V i nfxg for some i Ä n and x 2 X, then x is the only element of X such that A[fxg 2 X holds. But then there can only be one other set B with B [ fxg 2 X and B = V j nfxg hence a preference between A and B was already introduced by reverse strict independence. Analogously in the cases A = V i nfx; yg and A = V i nfx; y; zg for i Ä n and x ¤ y ¤ z 2 X, every possible instance of strict independence is already decided by dominance and two applications of strict independence.
It can easily be seen, that this construction does not lead to circles, if we start with a positive instance of betweenness: Every set of the form A = V i nfx; y; zg is only comparable to other sets by maximal dominance. Every set of the form A = V i nfx; yg is only comparable by maximal dominance or to another set of the same form. The order on sets of this form mirrors the order on sets of the form A = V i nfx; y; zg which is produced by maximal dominance and hence is circle free. Finally sets of the form A = V i nfxg or A = V i are only comparable to other sets by maximal dominance or if this is intended by the construction. Hence, the order on these sets is circle free, if we started with a positive instance of betweenness.
Finally, we can extend this order to a total order because extensions do not produce new instances of strict independence.

The Dominance Strict Independence problem
Proof We construct an instance .X; <; X / of the Dominance Strict Independence problem in a similar fashion as above. We take the same X and < and add the same sets to X . In order to make the reduction work for the Dominance Strict Independence problem, we have to add more sets.
Observe that maximal dominance is only needed in the reduction for the Maximal Dominance Strict Independence problem to introduce preferences like Anf1; k; k + 4g Bnf1; k + 1; k + 4g. We can enforce these preferences also using strict independence and regular dominance using a construction as in the proof of Proposition 1.
We want to enforce X k Y k . By definition, max.X k / = max.Y k / and min.X k / < min.Y k / or max.X k / < max.Y k / and min.X k / = min.Y k /. Assume, w.l.o.g. max.X k / = max.Y k / and min.X k / < min.Y k / and let X k = fx 1 ; x 2 ; :::; x l g and Y k = fy 1 ; y 2 ; :::; y m g be enumerations of X k resp. Y k such that i < j implies x i < x j resp. y i < y j . We add fx l g; fx l−1 ; x l g; :::; fx 2 ; :::; x l g and fx 1 ; x l g to X . This forces by dominance fx 2 ; :::x l g ::: fx l−1 ; x l g fx l g and hence by transitivity and strict independence X k fx 1 ; x l g. Analogously, we can enforce fx 1 ; y 1 ; y m g fx 1 g [ Y k by adding fx 1 ; y 1 g; fx 1 ; y 1 ; y 2 g; :::; fx 1 ; y 1 ; :::; y m−1 g as well as fx 1 ; y 1 ; y m g and fx 1 g [ Y k to X . Finally we add fx 1 ; y 1 ; y m g, fx 1 g and fx 1 ; y 1 g enforcing fx 1 g fx 1 ; y 1 g by dominance and hence fx 1 ; y m g fx 1 ; y 1 ; y m g by strict independence. Then, we have X k Y k by X k fx 1 ; x l g fx 1 ; y 1 ; x l g fx 1 g [ Y k Y k : The process of producing a positive instance of betweenness from a positive instance of the Dominance Strict In-K Fig. 5 Sketch of the sets V 1 ; :::; V n ; C 1 ; :::; F 1 ; F n 2 ; z and z Fig. 6 Enforcing strictness without strict independence dependence problem is the same as for the Maximal Dominance Strict Independence case. However, in order to construct a total order on X , we have to do a bit more. We take the same steps as in the Maximal Dominance Strict Independence case (including the closure under maximal dominance) but additionally, for A; B 2 X with max.A/ = max.B/, min.A/ = min.B/ and jAj; jBj Ä 3 we set A B if 1. min.A/ = 1 and jAj = 2 and jBj = 3, 2. min.A/ = N and jAj = 3 and jBj = 2, 3. jAj = jBj = 3 and AnB < BnA.
This order, together with a positive instance of betweenness, maximal dominance and (reverse) strict independence is circle free and decides all possible applications of strict independence. Therefore, we can construct a total order on X satisfying strict independence and dominance.

The Maximal Dominance Independence Problem
Proof We have to adapt the reduction for the Maximal Dominance Strict Independence problem above in two places. We have to change the way we enforce the strict preferences in Figure 1 and Figure 2 and we have to make sure that the order restricted to the sets V 1 ; V 2 ; :::; V n is strict.
To enforce, without strict independence, a strict preference between two sets that is not forced by maximal dominance we define for every pair A; B 2 X with min.B/ Ä min.A/, max.A/ Ä max.B/ and 2 Ä max.A/ − min.A/ a family of sets S.A; B/ forcing A B. S.A; B/ contains the following sets fx AB g; fy AB g; fx AB ; z AB g; fy AB ; z AB g; where min.A/ < y AB < x AB < max.A/, max.B/ < z AB and z AB < min.B/ hold.
Then A [ fz AB g fx AB ; z AB g holds by maximal dominance and, therefore, fx AB g 6 A and hence A fx AB g holds by "reverse" independence 6 and analogously, fy AB g B. Therefore, transitivity implies A B by A fx AB g fy AB g B.
For a triple .a; b; c/ in the instance of betweenness we add the following sets as in the two previous reductions: Anfkg; Bnfkg; Bnfk + 1g; C nfk + 1g; Anfk + 2g; Bnfk + 2g; Bnfk + 3g; C nfk + 3g 6 Every total order that satisfies independence also satisfies "reverse" independence, i. e., A B implies Anfxg Bnfxg. Otherwise, we would get Bnfxg Anfxg by the totality of , hence B A by independence, contradicting the definition of .
where we start with k = 4n 3 + 11n 2 + n + 7. We force the same preference as in the Maximal Dominance Strict Independence case, by adding the following families: S.Anfkg; Bnfk + 1g/; S.C nfk + 1g; Bnfkg/; S.Bnfk + 3g; Anfk + 2g/; S.Bnfk + 2g; C nfk + 3g/: We still have to make sure that the order on the sets V 1 :::V n is strict. To achieve this we want to use the idea shown in Figure 6, that is to add for every pair V i ; V j sets that lead to a circle if both V i V j and V j V i hold. Let f .l/ = .V i ; V j / be an enumeration of all pairs of sets V 1 ; V 2 ; :::; V n . We add sets C l ; D l ; E l and F l that are contained in the "middle parts" of all sets V i such that C l F l 0 holds for all l 0 < l. Moreover, we want the following: F l = E l nfmax.E l /; max 0 .E l /; min 0 .E l /; min.E l /g; E l = D l nfmax.D l /; max 0 .D l /; min 0 .D l /; min.D l /g; D l = C l nfmax.C l /; max 0 .C l /; min 0 .C l /; min.C l /g where max 0 .X/ denotes the second largest element of X and min 0 .X/ the second smallest. We can achieve this by taking for all l Ä n 2 C l W= f4n 3 + 4n 2 + n + 7l; :::; 16n 3 + 24n 2 + n + 14 − 7lg and D l ; E l and F l accordingly. Furthermore, we add sets F l nfmin.F l /g, V i [ fz l g and .F l nfmin.F l /g/ [ fz l g for a unique 7 z l < min.V 1 /. This ensures F l F l nfmin.F l /g V i . In a similar fashion we can enforce V i D l , E l V j and V j F l . Then, we add sets C l nfy l g, D l nfx l g, E l nfy l g and F l nfy l g for x l = 5n 3 + 11n 2 + n + 7 + l and y l = 5n 3 + 12n 2 +n+7+l. Furthermore, we enforce D l nfx l g F l nfy l g and E l nfy l g C l nfx l g by adding S.D l nfx l g; F l nfy l g/ and S.E l nfy l g; C l nfx l g/.
This forces a strict preference between V i and V j . Assume otherwise V i V j for a total order satisfying maximal dominance and independence. Then, for l such that f .l/ = .
However, then C l nfxg D l nfyg F l nfyg E l nfyg C l nfxg is a circle in , contradicting the assumption that is a total order.
It is straightforward to check that this construction yields a valid reduction analogously to the proof above. The key step is to observe that independence can only be applied to the new sets in cases where it is used in the proof. This is clear for the sets pictured in Figure 6. For the one and two element sets this holds, because the elements are unique and because no three element sets are contained in X .
It remains to check that we can actually pick a unique element every time we want to do this in the reduction. The inner part of F n 2 has .16n 3 +24n 2 + n +14−7n 2 −7/ − .4n 3 + 4n 2 +n+7n 2 +7/ = 12n 3 +6n 2 elements. For every triple, we have to pick 12 unique elements contained in this middle part. These are at most 12n 3 elements. Furthermore, we have to pick for every pair .V i ; V j / 6, two for x and y and 4 to enforce preferences, unique elements. There are n 2 such pairs, so we need 6n 2 elements. Hence we need at most the 12n 3 +6n 2 elements contained in F n 2 . Furthermore, we need for every pair .V i ; V j / and every triple 4 elements smaller than min.V 1 / and the same amount of elements larger than max.V 1 /. This is possible as min.V 1 / = 4n 3 + 4n 2 + 1 and N − max.V 1 / = .20n 3 + 28n 2 + 2n + 14/ − .16n 3 + 24n 2 + 2n + 14/ = 4n 3 + 4n 2 .

The Dominance Independence problem
case was adapted to the Dominance Strict Independence case.

Conclusion
We have shown that the problem of deciding whether a linear order can be lifted to a ranking of sets of objects satisfying a form of dominance and a form of independence is in P or trivial if we do not expect the ranking to be total and NP-complete if we do. In order to prove P-membership or triviality we constructed such rankings. Rankings of specific sets are useful in several applications, e.g., to eliminate obviously inferior sets of objects from a set of options.
In many applications, the family of sets to be ranked is not given explicitly but implicitly. We expect that a compact representation of the sets increases the computational complexity of decision problems as studied in this paper. As future work we thus want to investigate the complexity blow up caused by a compact representation.
Furthermore, we would like to characterize families that allow an order satisfying (maximal) dominance and (strict) independence. Moreover, it may be possible to find sufficient but not necessary conditions for the existence of such rankings that can be checked in polynomial time. We aim for finding strong forms of such conditions. A related goal is to obtain special classes of families where such a characterization is feasible. A promising candidate are families generated via graphs, where the family is given by the sets of vertices that induce connected subgraphs.
Another item on our agenda is to investigate whether the logic proposed in [8] can be used for specific sets of objects as well. Finally, it would be interesting to study some of the other axioms that have been considered in the literature and see how they behave when one has to rank proper subsets of the whole power set of elements.