Determining a Slater Winner is Complete for Parallel Access to NP

03/30/2021 ∙ by Michael Lampis, et al. ∙ Université Paris-Dauphine 0

We consider the complexity of deciding the winner of an election under the Slater rule. In this setting we are given a tournament T = (V, A), where the vertices of V represent candidates and the direction of each arc indicates which of the two endpoints is preferable for the majority of voters. The Slater score of a vertex v∈ V is defined as the minimum number of arcs that need to be reversed so that T becomes acyclic and v becomes the winner. We say that v is a Slater winner in T if v has minimum Slater score in T. Deciding if a vertex is a Slater winner in a tournament has long been known to be NP-hard. However, the best known complexity upper bound for this problem is the class Θ_2^p, which corresponds to polynomial-time Turing machines with parallel access to an NP oracle. In this paper we close this gap by showing that the problem is Θ_2^p-complete, and that this hardness applies to instances constructible by aggregating the preferences of 7 voters.

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1 Introduction

We investigate the computational complexity of the following voting scheme, often referred to as the Slater rule. We are given a set of candidates and the preferences of voters, where each voter’s preferences correspond to her own complete ranking of . We determine a pair-wise relation on as follows: for we say that wins against if the majority of voters prefers candidate over candidate

. In this way, assuming that there are no ties (which is guaranteed if the number of voters is odd), we can construct a tournament

, where we have the arc (that is ) if wins against . In this setting, the Slater score of a candidate is the minimum number of arcs of that need to be reversed so that becomes transitive (acyclic) with being placed last (that is, with being a sink). The Slater winner of a tournament is a candidate with minimum Slater score. Intuitively, a candidate is a Slater winner if there exists a linear ordering of the candidates that ranks as the winner and is as compatible as possible with the voters’ aggregated preferences, in the sense that the edit distance between and is minimum.

The notion of Slater winner is very natural and can be seen as a special case of Kemeny voting. Indeed, in Kemeny voting we construct a weighted tournament where the weight of the arc denotes the margin of victory of over . In this sense, the Slater system corresponds to a version of Kemeny voting where we only retain as information which of the two candidates would win a head-to-head match-up. In other words, Slater voting is the special case of Kemeny voting where the arcs are unweighted. For more information about these and other related voting systems, we refer the reader to [5].

The main question we are interested in in this paper is the computational complexity of determining if a vertex of a tournament is a Slater winner. It has long been known that this question is at least NP-hard [14]. Indeed, it is not hard to see that if we had an oracle for the Slater problem we would be able to produce in polynomial time an ordering of any tournament in a way that minimizes the number of inversed arcs. This would solve the Feedback Arc Set problem, which is known to be NP-complete on tournaments [1, 2, 6, 8]. On the other hand, membership of this problem in NP is not obvious. The best currently known upper bound on its complexity is the class , shown by Hudry [7, 14].

The class seems like a natural home for the Slater problem, as this class captures as a model of computation polynomial-time Turing machines which are allowed to use an NP oracle either a polynomial number of times non-adaptively (that is, with questions not being allowed to depend on previous answers), or a logarithmic number of times adaptively. Intuitively, solving the Slater problem requires us to calculate exactly a value that is NP-hard to compute (the minimum feedback arc set of a tournament). This can be done either by asking polynomially many non-adaptive NP queries to an oracle (for each we ask if the feedback arc set has size at most ), or a logarithmic number of adaptive queries (where we essentially perform binary search). It has therefore been conjectured that determining if a candidate is a Slater winner is -complete [4, 5, 14].

The result we present in this paper confirms this conjecture and proves that the problem is indeed -complete. This places Slater voting in the same class as related voting schemes, such as Kemeny [13], Dodgson [11], and Young [15]. It also places it in the same class as the Slater rule used in [10] for a more general judgment aggregation problem. We prove this result by modifying the reduction of Conitzer [8], which showed that Feedback Arc Set on tournaments is NP-complete. The main difference is that, rather than reducing from SAT, we need to reduce from a -complete variant, where we are looking for a maximum weight satisfying assignment that sets a certain variable to True. This forces us to complicate the reduction so that we can encode in the objective function not only the number of satisfied clauses but also the weight of the corresponding assignment.

Having settled the worst-case complexity of the problem in general, we go on to consider a related question: what is the minimum number of voters for which determining a Slater winner is -complete? The motivation behind this question is that, even though any tournament can be constructed by aggregating the preferences of a large enough number of voters111This is a classical result known in the literature as McGarvey’s theorem., if the number of voters is limited, some tournaments can never arise. Hence the problem may conceivably be easier if the number of voters is bounded. In the case of the Slater rule, Bachmeier et al.[3] have shown that determining the Slater winner remains NP-hard for voters. By reusing and slightly adjusting their arguments we improve their complexity lower bound to -completeness for voters.

2 Definitions and Preliminaries

A tournament is a directed graph such that for all , exactly one of the arcs appears in . A feedback arc set (fas) of a digraph is a set of arcs such that deleting from results in an acyclic digraph. If is a tournament and is a fas of , then the tournament obtained from by reversing the direction of all arcs of is acyclic (or transitive). We will say that a total ordering of the vertices of a digraph implies the fas (in the sense that is the set of arcs that disagree with the ordering). We will say that an ordering of is optimal if the fas it implies has minimum size.

Given a digraph and , we say that is a Slater winner if for some the following hold: (i) there exists a fas of , such that is a sink of and (ii) every fas of has size at least . If is a Slater winner in , then a winning ordering for is a linear ordering of that places last and implies a fas of of minimum size.

In a digraph , a set is a module if the following holds: for all and we have and . In other words, every vertex outside that has an arc to (respectively from) a vertex of , has arcs to (respectively from) all of . The following lemma, given by Conitzer [8] with slightly different terminology, states that the vertices of a module can, without loss of generality, always be ordered together. We say that the vertices of a set are contiguous in an ordering if there are no , such that .

Let be a digraph, a vertex, and suppose we have a partition of into non-empty modules . If is a Slater winner of , then there exists a winning ordering for such that for all , the vertices of are contiguous.

Proof.

Suppose (otherwise the claim is trivial) and consider an ordering that is winning for . We will say that a set of vertices is a block of if (i) for some ; (ii) is contiguous; (iii) is maximal, that is, adding any vertex to violates one of the two preceding properties.

If the number of blocks is equal to we are done, as each block is equal to a module so we have an ordering where each module is contiguous. If we have at least blocks, we will explain how to edit the ordering so that it remains winning for , it implies a fas of the same size, and the number of blocks decreases. Repeating this process until we have blocks completes the proof.

Consider two vertices , with , which belong to the same module, say , but in distinct blocks. Among all such pairs, select so that their distance in the ordering, that is, the size of the set is minimized. Let be the blocks that contain respectively. Note that by the selection of we have that is the last vertex of , is the first vertex of , , and .

Let (respectively ) be the out-degree (respectively in-degree) of towards the set . Because is a module, all vertices of have the same in-degree and out-degree towards , and in particular, and . Now, if , we can obtain an ordering that implies a smaller fas by placing immediately after the last vertex of . This would contradict the optimality of , so it must be impossible. Similarly, if , we have , and we can obtain a strictly better ordering by placing immediately before the first vertex of , contradiction. We conclude that . Therefore, moving all the vertices of so that they appear immediately after the last vertex of produces an ordering which is equally good as the current one, is still winning for , and has a smaller number of blocks. ∎

2.1 Complexity

We recall the class which is known to have several equivalent characterizations, including (P with the right to make queries to an NP oracle), (logarithmic-space Turing machines with access to an NP oracle), and (P with parallel non-adaptive access to an NP oracle). We refer the reader to [12] for more information on this class. In [16] it was shown that the following problem is -complete: given a graph , is the maximum clique size odd? In [9] it is mentioned that the following problem, called Max Model, is -complete: given a satisfiable CNF formula containing a special variable , is there a satisfying assignment of that sets to True and has maximum Hamming weight (among all satisfying assignments), where the Hamming weight of an assignment is the number of variables it sets to True.

We will use as a starting point for our reduction a variant of Max Model which we show is -complete below.

The following problem is -complete. Given a 3-CNF formula containing a distinguished variable , such that is satisfied by the all-False assignment, decide if there exists a satisfying assignment for that sets to True and has maximum weight among all satisfying assignments.

Proof.

We start with a graph for which the question is if the maximum independent set has odd size (clearly this is equivalent to the question of deciding if the maximum clique has odd size by taking the complement of , so our starting problem is -complete). Let and suppose . We construct a formula as follows: for each we build variables and for each we add the clause ; for each , for each we add the clause ; we construct variables and clauses that represent the constraints , and for each , . We set as the distinguished variable of . The formula construction can clearly be carried out in polynomial time, and no clause has size more than three. Furthermore, setting everything to False satisfies all clauses. Intuitively, for each vertex we have constructed variables that will be set to True if we take this vertex in the independent set. The first set of clauses ensures that we make a consistent choice among the copies; the second set that we indeed select an independent set; and the third calculates the parity of its size.

We now observe that independent sets of naturally correspond to satisfying assignments of . In particular, given an independent set we can construct an assignment by setting, for all , to True if and only if ; we then complete the assignment by giving appropriate values to the variables so that the parity constraints are satisfied. For the converse direction, we can extract an independent set from a satisfying assignment by setting if and only if the assignment sets to True. We observe the is set to True in a satisfying assignment if and only if the corresponding independent set has odd size (indeed, for each , is set to True if the intersection of the independent set with the first vertices has odd size).

Suppose now that there exists an independent set of maximum size and that is odd. Then, there exists a satisfying assignment of maximum weight that sets to True. Indeed, suppose for contradiction that the maximum satisfying assignment sets to False. Then, the corresponding independent set must have even size . Since is odd and is a maximum independent set, . But then, the weight of is at most . However, the assignment corresponding to has weight at least , contradiction.

For the converse direction, suppose there exists a satisfying assignment of maximum weight that sets to True. The corresponding independent set has odd size, say . If there exists a maximum independent set that has even size , then . However, the corresponding truth assignment would have weight at least . Since has weight at most we get a contradiction to the optimality of .

We conclude that there is a satisfying assignment to of maximum weight that sets to True if and only if the maximum independent set of has odd size. ∎

3 Reduction to Slater

The following problem is -complete: given a tournament and a vertex , decide if is a Slater winner.

Proof.

We perform a reduction heavily inspired by the reduction of [8] proving that computing the minimum fas of a tournament is NP-complete, though we include a minor modification proposed by Bachmeier et al. [3] which will later allow us to show that our instances are realizable using seven voters. The main complication compared to the reductions of [8, 3] is that we now need to encode the CNF formula in a way that satisfying assignments of larger weight correspond to orderings with better objective value.

We start with a formula , as given in Lemma 2.1. Let be the variables of and suppose that the question is whether there exists a satisfying assignment of maximum weight that sets to True. Recall that by assumption the all-False assignment satisfies . Let be the number of clauses of .

We define two numbers which satisfy the following properties:

(1)
(2)
(3)

For concreteness, set and and the above inequalities are easily satisfied when is sufficiently large. Importantly, are polynomially bounded in . Intuitively, the idea is that are two very large numbers, and is significantly larger. We will construct modules of size (roughly) or , and the values are chosen so that arcs between large modules will be very important, arcs between large and small modules quite important, and arcs between small modules almost irrelevant.

We now construct our tournament as follows: for each we construct modules, call them . Modules have size , while modules have size if , and the size of is . Internally, each of these modules induces a transitive tournament. For we add all arcs from to . For each we add all possible arcs (i) from to (ii) from to (iii) from to (iv) from to (v) from to (vi) from to . The graph we have constructed so far is a tournament with sections, each made up of modules. Each such section represents a variable and the sections are linearly ordered. The structure inside each section is essentially the transitive closure of with the exception that arcs between and are heading towards .

We now complete the construction by adding to the current tournament some vertices that represent the clauses of . In particular, for each we construct a module of size to represent the -th clause of . Internally, is a transitive tournament. For , the arcs between and are set in an arbitrary direction. What remains is to explain how the arcs between and the modules representing the variables are set so as to encode the incidence of variables with clauses. For each and we do the following:

  1. If does not appear in the -th clause we add all arcs from to and all arcs from to .

  2. If appears positive in the -th clause we add all arcs from to and all arcs from .

  3. If appears negative in the -th clause we add all arcs from to and all arcs from .

This completes the construction and the question we want to answer is whether the last vertex (that is, the sink) of the transitive tournament induced by is a Slater winner of the whole graph.

We need to prove that the designated vertex is a Slater winner if and only if there is a satisfying assignment for with maximum weight that sets to True. We will do this by establishing some properties regarding any optimal ordering of the constructed tournament, showing that such an ordering must always have a structure which implies a satisfying assignment of with maximum weight. We will rely heavily on Lemma 2, since the tournament we have constructed can be decomposed into modules, namely, , and , for , and , for . We therefore assume without loss of generality that these sets are placed contiguously in an optimal ordering.

Let us first argue that any optimal ordering must have some desirable structure which necessarily encodes a satisfying assignment for . To do this it will be helpful to start with a baseline ordering and calculate its implied fas, as then any ordering which implies a larger fas will be necessarily suboptimal. Consider the ordering which is defined as for each and which sets , where each module is internally ordered in the optimal way. We insert into this ordering of the modules that represent variables, the modules as follows: for each , we find a variable that appears negative in the -th clause (such a variable must exist, since is satisfied by the all-False assignment), and place all of between and . If for some pair their relative ordering is not yet fully specified, we order them in some arbitrary way.

The arcs incompatible with the above ordering are (i) the at most arcs going from a module to a module (ii) for each that was placed between and we have arcs (towards ), as well as at most arcs to each other group , for (iii) the total number of arcs between modules is at most . Therefore, we have that the fas implied by this ordering has size at most

In the remainder we will therefore only consider orderings which imply a fas of size at most , as other orderings are suboptimal. This allows us to draw some conclusions regarding the structure of an optimal ordering. First, observe that for each , any ordering of will contribute at least arcs to the fas. Using inequality (1), we have that there are at most pairs of “large” modules (that is, modules of size at least ) which are incorrectly ordered, that is, ordered so that all arcs between the modules are included in the fas. Indeed, if there are such pairs, the fas will have size at least . We conclude that regarding the large modules we must have the following ordering:

  1. For each , we have that all vertices of (the section that represents the variable ) are before all vertices of (the section that represents the variable ).

  2. For each , we have and all vertices of are before .

  3. For each we have , or , or .

We would now like to construct a correspondence between assignments to and orderings of the tournament that respect the above conditions. On the one hand, if we are given an assignment we construct an ordering of the variable sections as above and for each , if set to True we set , otherwise we set . In the converse direction, given an ordering that respects the above conditions (which any optimal ordering must do), we extract an assignment by setting, for each , to True if and only if .

We now argue that the assignment corresponding to an optimal ordering must also be satisfying for , as otherwise the fas will have size strictly larger than , contradicting the optimality of the ordering. For the sake of contradiction, suppose we have an optimal ordering which corresponds to an assignment falsifying a clause. As argued above, there are at least arcs in the fas contributed by the ordering of the large modules, so we concentrate on the modules representing clauses. A module representing any clause must be incident on at least arcs of the fas connecting it to large modules. To see this, consider the following: we will say that is in the interior of section , if and is placed before one of , or . can be in the interior of at most one section , so for each we observe that at least arcs incident on and modules of the group are in the fas. This gives arcs. In addition, no matter where we place in the interior of section , at least a further arcs of the fas are obtained: if is after , then we get the arcs to or the arcs to ; if is between and , we get the arcs to and at least arcs from . Hence, we get at least arcs in the fas for each .

Furthermore, suppose that the assignment corresponding to the ordering does not satisfy the -th clause. Then, we claim that at least arcs connecting to large modules are included in the fas. Indeed, if is in the interior of section , it can either be before or after . If it is before , as we observed in the previous paragraph, we always have at least arcs in the fas between and the large modules of section . If is placed after , we have the following cases: (i) if does not appear in the clause, then at least arcs between and the large modules of section are in the fas (ii) if appears positive in the -th clause, we know that is not placed last in section (otherwise the assignment would satisfy the -th clause), so wherever we place , at least arcs are included in the fas (iii) similarly if appears negative, since the assignment does not satisfy the clause, is last, so again at least arcs are included in the fas.

From the above calculations, if the assignment that corresponds to an ordering falsifies a clause, the fas has size at least , where we used inequality (2). We conclude that an optimal ordering must correspond to a satisfying assignment.

We now need to argue that the assignment corresponding to an optimal ordering of the tournament must be a satisfying assignment of maximum weight. Suppose for contradiction that the assignment corresponding to an optimal ordering, call it , sets variables to True, but there exists another satisfying assignment, call it , that sets at least variables to True. We will show that starting from we can obtain a better ordering of the tournament, contradicting the optimality of the original ordering.

We claim that the ordering from which we extracted includes at least arcs in the fas. This is because in the section corresponding to the variables that sets to False, either the arcs from to , or the arcs from to are in the fas (since is not placed last in the section), and these are at least arcs.

We construct an ordering from as follows: we order the variable section in the normal way and inside each section, if we use the ordering , otherwise we use the ordering . For each , we find a variable that satisfies the -th clause and place in section immediately before the last module of this section. If for the order of is not implied by the above, we set it arbitrarily. The fas implied by this ordering has size at most

Here, the calculations for the terms are the same as in the calculation of ; the term takes into account that there are sections that correspond to variables set to False; and the term is due to the fact that may be one of the variables set to False and has size and not . Using inequality (3) we have that , so the ordering we have constructed from is better than the one from which we extracted , contradiction.

At this point we are almost done because we have argued that an optimal ordering of the tournament corresponds to a satisfying assignment of maximum weight and furthermore, since the correspondence sets to True if and only if is the last module in the ordering, the sink of will be last in the ordering if and only if the assignment sets to True. However, could have several satisfying assignments of the same weight, and since we have set arcs between modules arbitrarily, it could be the case that a maximum weight assignment that sets to False results in a better ordering, making another vertex the Slater winner. This is the reason why we have set to be slightly larger than all other modules , so that setting to True is always slightly more advantageous than setting any other variable to True.

Concretely, we argue the following: any optimal ordering of the tournament corresponds to a satisfying assignment of with maximum weight; and furthermore if a satisfying assignment of with maximum weight sets to True, then any optimal ordering places last. We need to argue the second claim, so suppose for contradiction that an optimal ordering does not place last and that the assignment that corresponds to this ordering is . Furthermore, suppose that there exists a satisfying assignment of maximum weight that sets to True. Say that both set variables to True.

We first observe that the ordering from which we extracted implies a fas of size at least . This is because there are sections where the fas contains arcs incident on a module , sections where the fas contains arcs incident from to , and in the section corresponding to the fas contains arcs, incident on .

On the other hand, if we construct an ordering from in the same way as we did previously, the fas obtained will have size at most

Again, using inequality (3) which states that we conclude that the new ordering is better, contradicting the optimality of the original ordering.

We now summarize our arguments: we have shown that any optimal ordering of the tournament always corresponds to a maximum weight satisfying assignmet of and furthermore, it corresponds to a maximum weight satisfying assignment that sets to True if this is possible; furthermore, if an optimal ordering corresponds to an assignment that sets to True then the last vertex of is a Slater winner. We therefore have two cases: if the last vertex of is a Slater winner, then since optimal orderings give rise to satisfying assignments of maximum weight, there is a maximum weight satisfying assignment of setting to True; if the last vertex of is not a Slater winner, then the maximum weight satisfying assignment we extract from an optimal ordering sets to False, and there is no satisfying assignment of the same weight setting to True. We conclude that determining if a vertex is a Slater winner is equivalent to deciding if has a maximum weight satisfying assignment setting to True, and is therefore -complete. ∎

4 Hardness for Voters

In this section we show that the tournaments constructed in Theorem 3 correspond to instances that could result from the aggregation of the preferences of voters and as a result the problem of determining a Slater winner remains -complete even for voters. Our approach follows along the lines of the arguments of Bachmeier et al. [3] who proved that determining the Slater winner is NP-hard for voters. Indeed, the proof of [3] consists of an analysis (and tweak) of the construction of Conitzer [8] which establishes that the instances of the reduction can be built by aggregating voter profiles. Since our reduction is very similar to Conitzer’s, we essentially only need to adjust the arguments of Bachmeier et al. to obtain -completeness.

Our first step is to slightly restrict the -complete problem that is the starting point of our reduction. We present the following strengthening of Lemma 2.1, which is similar to the problem used as a starting point in the reduction of [3].

The problem given in Lemma 2.1 remains -complete under the following additional restrictions: (i) we are given a partition of the clauses of in two sets and each variable appears in at most one clause of and in at most two clauses of (ii) each literal appears at most once in a clause of .

Proof.

Given a formula as in Lemma 2.1 we construct a new formula as follows. Let be the variables of and be the number of its clauses. For each , we construct variables, call them . For each , for we construct the clause , as well as the clause . Let be the set of clauses constructed so far and note that each literal appears at most once and each variable at most twice in these clauses. Intuitively, the clauses of ensure that for each , all variables in the set must receive the same value in a satisfying assignment.

Now, we consider the clauses of one by one. If the -th clause contains the variable , we replace it by the variable . Doing this for all clauses of we obtain a set of clauses, call it , where each variable appears at most once (assuming without loss of generality that clauses of have no repeated literals).

If was the designated variable of we set as the designated variable of . It is now not hard to make a correspondence between satisfying assignments of and ( is set to True if all are set to True) in a way that preserves weights (the weight of an assignment to is times the weight of the corresponding assignment for ). Hence, determining if a maximum weight satisfying assignment to sets to True is -complete. Observe also that is satisfied by the all-False assignment. ∎

We now obtain the result of this section by starting the reduction of Theorem 3 from the problem of Lemma 4. In the statement of the theorem below, when we say that a tournament can be obtained from voters, we mean that there exist total orderings of such that for all we have that in at least of the orderings.

Determining if a vertex of a tournament is a Slater winner remains -complete even for tournaments that can be obtained from voters.

Proof.

We perform the same reduction as in Theorem 3 except we start from the special case given in Lemma 4. What remains is to show that the instance we construct can result from aggregating orderings. Recall that our tournament contains modules representing the variables, called , for and modules representing the clauses, called , for . Since modules are internally transitive, we will assume that the voters have preferences which agree with the directions of the arcs inside the modules and hence we focus on the arcs between modules. Recall that in the reduction of Theorem 3, arcs between modules are set arbitrarily. To ease presentation, assume that when we have the arcs .

The first voter has preferences . In other words, the first voter orders all the variable modules before all the clause modules, orders variable groups according to their index, and inside each variable group she has the ordering .

We now add two voters with the intent of constucting all the arcs of the set

The first of these voters has ordering . The second of these voters has ordering . Note that these two voters agree that all modules of come before all modules of and that for each we have , but disagree on every other pair of modules, hence the two voters together induce exactly the set of arcs cited above.

If we now consider the three voters we have so far we observe that much of our construction is already induced:

  1. For we have arcs from to because of the preferences of the first voter, as the other two voters disagree on these arcs.

  2. For each , inside the group , we have arcs that agree with the ordering , except that we have arcs from to . This is because the second and third voter agree that , but disagree on every other pair (hence the preferences of the first voter prevail for the other pairs).

  3. For each we have arcs from to , due to the preferences of the first voter, as the other two disagree.

  4. For each and we have arcs from to , because the second and third voter agree that (though the first voter disagrees).

  5. For each and we have arcs from to , because the second and third voter disagree on these pairs, so the preferences of the first voter break the tie.

We therefore have that the tournament that follows from aggregating the preferences of the first three voters almost corresponds to the one we want to construct, except that for each and the arcs between and correspond to the arcs we would want if did not appear in the -th clause (in other words, the three voters we have so far induce the general structure of the construction, but do not encode which variable appears in which clause). Furthermore, if we look at the relationship between any two modules so far, the margin of victory is always exactly one (that is, there do not exist two modules such that all three voters agree that ).

Hence, what remains is to use the four remaining voters to “fix” this, so that if appears (positive or negative) in the -th clause, we have the arcs prescribed in the reduction of Theorem 3. We will achieve this by giving two pairs of voters. Each pair of voters will disagree on all pairs of modules except a specific set of arcs that we want to fix. Hence, adding the pair of voters to the electorate will repair the arcs in question (since the current margin of victory for all arcs is one), while leaving everything else unchanged.

Recall that the clause set is given to us partitioned into two sets so that each variable appears in at most one clause of and each literal in at most one clause of . We will use the slightly weaker property that each literal appears at most once in each of . Abusing notation we will write if the -th clause is in (similarly for ). We will also write (respectively ) if appears positive (respectively negative) in the -th clause.

Consider now the following two sets of arcs:

Our plan is to give a pair of voters whose preferences induce the arcs of and another pair whose preferences induce the arcs of . Here when we say that two voters induce a set of arcs we mean that for each both voters have and for each one voter has and the other has . Before we proceed we observe that if are inducible by a pair of voters each, then adding these four voters to the three voters we have described so far produces the tournament of Theorem 3. Indeed, suppose that appears positive in clause and . Then, if we consider the arcs in the tournament induced by the first three voters, we need to inverse the arcs between and (which currently point ), and the arcs between and (which currently point ). But the arcs between and are inversed thanks to , while the arcs between and are inversed thanks to , where we use the fact that represent the consensus of two voters, while the margin of victory for any arc induced by the first three voters is one. Similar arguments apply if appears negative in , or . Hence, if a pair of voters induces and another induces , taking the union of these four voters with the three voters we have described produces the tournament of Theorem 3 and completes the proof.

We now recall that it was shown in [3] that are inducible by two voters each, since these sets of arcs are unions of stars (if we contract each module to a vertex). Let us explain in more detail how to represent as the union of the preferences of two voters (the arguments for are essentially identical). We will make use of the fact that each literal appears at most once in and at most once in .

We will say that a module from a variable group is “active” if it is incident on an arc of . In particular, modules , for are not active, and neither are modules such that does not appear positive in (and similarly for ). We will concentration on the ordering of active modules because if we find two voter profiles that order these modules in a way that induces , we can add an arbitrary ordering of the inactive modules in the beginning of the preferences of the first voter, and the opposite of that ordering at the end of the preferences of the second voter. This will have as effect that the two voters disagree on any pair that involves an element of an inactive module, as desired.

We now observe that for each active module from , there exists exactly one such that has arcs to in . This is because every literal appears in at most one clause of and at most one clause of . Now, we construct two voter profiles as follows: one voter orders the modules in increasing order of index and the other in decreasing order. For each active module or , we insert the module immediately after the from which the module receives arcs in in both orderings; for active modules or we insert them immediately before the towards which the module has arcs in both orderings. Note that this does not fully specify the ordering, as if two variables appear in and , then we need to place and immediately after . We resolve such conflicts by using an arbitrary ordering of the active modules for the first voter and the opposite of that ordering for the second voter, that is, all modules which are supposed to appear immediately after are sorted in one way for the first voter and in the opposite way for the second voter. We now observe that with this ordering for every active module the two voters agree about the arcs connecting the module to its neighboring , while we obtain no other arcs between the module and any other or any other active module. We therefore have two voters whose preferences induce . ∎

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