# The Complexity of the Possible Winner Problem over Partitioned Preferences

The Possible-Winner problem asks, given an election where the voters' preferences over the set of candidates is partially specified, whether a distinguished candidate can become a winner. In this work, we consider the computational complexity of Possible-Winner under the assumption that the voter preferences are partitioned. That is, we assume that every voter provides a complete order over sets of incomparable candidates (e.g., candidates are ranked by their level of education). We consider elections with partitioned profiles over positional scoring rules, with an unbounded number of candidates, and unweighted voters. Our first result is a polynomial time algorithm for voting rules with 2 distinct values, which include the well-known k-approval voting rule. We then go on to prove NP-hardness for a class of rules that contain all voting rules that produce scoring vectors with at least 4 distinct values.

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## 1 Introduction

In political elections, web site rankings, and multiagent systems, preferences of different parties (voters) have to be aggregated to form a joint decision. A general solution to this problem is to have the agents vote over the alternatives. The voting process is conducted as follows: each agent provides a ranking of the possible alternatives (candidates). Then, a voting rule takes these rankings as input and produces a set of chosen alternatives (winners) as output. However, in many real-life settings one has to deal with partial votes: Some voters may have preferences over only a subset of the candidates. The Possible-Winner problem, introduced by Konczak and Lang [11] is defined as follows: Given a partial order for each of the voters, can a distinguished candidate win for at least one extension of the partial orders to linear ones ?

The answer to the Possible-Winner problem depends on the voting rule that is used. In this work we consider positional scoring rules. A positional scoring rule provides a score value for every position that a candidate may take within a linear order, given as a scoring vector of length in the case of candidates. The scores of the candidates are added over all votes and the candidates with the maximal score win. For example, the -approval voting rule, typically used in political elections, defined by starting with ones, enables voters to express their preference for candidates. Two popular special cases of -approval are plurality, defined by , and veto, defined by .

The Possible-Winner problem has been investigated for many types of voting systems [4, 13, 18, 21]. For positional scoring rules, Betzler and Dorn [3] proved a result that was just one step away from a full dichotomy for the Possible-Winner problem with positional scoring rules, unweighted votes, and any number of candidates. In particular, they showed NP-completeness for all but three scoring rules, namely plurality, veto, and the rule with the scoring vector . For plurality and veto, they showed that the problem is solvable in polynomial time, but the complexity of Possible-Winner remained open for the scoring rule until it was shown to be NP-complete as well by Baumeister and Rothe [2].

Partitioned preferences provide a good compromise between complete orders and arbitrary partial orders. Intuitively, the user provides a complete order over sets

of incomparable items. In the machine learning community, partitioned preferences were shown to be common in many real-life datasets, and have been used for learning statistical models on full and partial rankings

[14, 17, 10].

In many scenarios, the user preferences are inherently partitioned. In recommender systems, the items are often partitioned according to their numerical level of desirability [19] (e.g., the common star-rating system, where the scores range between and stars). In such a scenario, all items with identical scores are incomparable. In some e-commerce systems, user preferences are obtained by tracking the various actions users perform [12]. For example, searching or browsing a product is indicative of weak interest. Bookmarking it is indicative of stronger interest, followed by entering the product to the “shopping cart”. Finally, the strongest indication would be actually purchasing the product. In this case as well, the items are partitioned into groups, where the desirability of each group is determined by its set of associated actions, and items in a common group are considered incomparable. In the field of information retrieval, learning to rank [5, 16] refers to the process of applying machine learning techniques to rank a set of documents according to their relevance to a given query. In this setting, document scores are indicative of relevance to the query, and documents with identical scores are considered incomparable.

In this work we investigate the computational complexity of the Possible-Winner problem with partitioned preference profiles. Our first result is that determining the possible winner can be performed in polynomial time for -valued voting rules (i.e., that produce scoring vectors with distinct values), which include the -approval voting rule. We then show that our algorithm also solves the possible winner problem for the voting rule. These result are surprising because both of these rules are NP-complete when the partitioned assumption is dropped [3, 2]. We then go on and prove hardness for the class of voting rules that produce scoring vectors containing at least distinct values, and a large class of voting rules with distinct values. The hardness proofs are involved because many of the order restrictions applied in the reductions for the general case are unavailable under the constraint of partitioned preferences.

## 2 Preliminaries

In this section we present some basic notation and terminology that we use throughout the manuscript.

### 2.1 Orders and rankings

A partially ordered set is a binary relation over a set of alternatives, or candidates that satisfies transitivity ( and implies ) and irreflexivity ( never holds). A linear (or total) order is a partially ordered set where every two items are comparable. We say that a total order extends the partial order if, for every pair of alternatives, such that it also holds that . We denote by the set of all linear orders over , and by the set of linear orders over that extend . In this manuscript we consider a special type of partial order termed partitioned preferences. [Partitioned preferences [15]] A partial order is a partitioned preference if the set of candidates can be partitioned into disjoint subsets such that: (1) for all , if and then ; and (2) for each , candidates in are incomparable under (i.e., and for every ).

### 2.2 Elections

Let be a set of voters, and a set of candidates. Every voter has a preference, also denoted , which is a linear order or complete vote over (i.e., ). A tuple of complete votes is an -voter preference profile. The set of all preference profiles on is denoted by . A voting rule is a function from the set of all profiles on to the set of nonempty subsets of . Formally . For a voting rule , and a preference profile , we say that candidate wins the election (or just wins) if , and co-wins if . We denote an election by the triple .

We now generalize the election to the case where some or all of the votes are partial orders over the candidates. We consider the election where the voter profile is comprised of partial orders over the candidates. We say that a profile extends the profile if they have the same cardinality (i.e., ), and every vote is a linear order that extends the partial order (i.e., ). We say that a partial preference profile is partitioned if every one of its preferences is partitioned. [-possible winner (co-winner)] Given an election where is a profile of partial orders over the candidate set , and a distinguished candidate , does there exist an extension of such that ()?

### 2.3 Positional scoring rules

Let denote an election with candidates and voters. A positional scoring rule is defined by a sequence of -dimensional scoring vectors where are positive integers denoted score values, and for every . A voting rule is normalized if for every there is no integer greater than one that divides all score values in , and . Since these assumptions have been shown to be non-restrictive [9, 3] we will consider only normalized scoring vectors in this work. We say that a positional scoring rule is pure [3, 7] if for every , the scoring vector for candidates can be obtained from the scoring vector for candidates by inserting an additional score value at an arbitrary position such that the resulting vector meets the monotonicity constraint. We note that for voting rules that are defined for a constant number of candidates, the possible winner problem can be decided in polynomial time [6, 20].

Given a complete vote , and a candidate , we define the score of in by where is the position of in . The score of candidate in a profile is defined as . Whenever the profile is clear from the context, we write . A positional scoring rule selects as winners all candidates with the maximum score .

Some popular examples of positional scoring rules are Borda, for which the scoring vector is , plurality, for which the scoring vector is , veto, for which the scoring vector is , and -approval , for which the scoring vector is . We assume that the scoring vector, and thus the scores of the candidates, can be computed in polynomial time given a complete profile.

## 3 Summary of Results

In this manuscript we consider the possible winner problem over partitioned preferences (Definition 2.1). We assume that all positional scoring rules are normalized.

[-valued voting rule] We say that a positional scoring rule is -valued if there exists a number such that for all , the score vector contains exactly distinct values. By this definition, the -approval, veto, and plurality voting rules are -valued, while Borda has an unbounded number of different score values.

[unbounded-value voting rule] We say that a positional scoring rule has an unbounded number of positions with equal score values if, for every , there exists a number such that for all , the score vector contains at least consecutive positions where .

Let denote an election where is a set of candidates, is a positional scoring rule, and is a partial profile where all of the votes are partitioned. In the rest of the manuscript we show the following. If is -valued, or if is then we show that the Possible-Winner problem over can be solved in polynomial time. In particular, this means that the Possible-Winner problem is tractable for the -approval voting rule. This result is surprising because it has been shown that when the partitioned assumption is dropped, the problem is intractable for both -approval [3], and  [2].

Our hardness results, proved in Section 5, cover all scoring rules that produce scoring vectors with at least distinct values. For -valued scoring rules, we prove hardness for all rules except vectors of the form where and are fixed constants such that , for which the complexity remains open. The main results are summarized in Theorem 3. A scoring rule is called differentiating [7] if it produces a scoring vector that contains two positions where such that .

Let be a positional scoring rule. Then we have the following when the preference profile is partitioned.

1. If is -valued or if is , then the Possible-Winner problem over can be answered in polynomial time.

2. If produces a scoring vector with at least distinct values then the Possible-Winner problem is NP-complete for .

3. If is -valued, and produces a size- scoring vector that is differentiating, or where the number of positions occupied by either or is unbounded, then the Possible-Winner problem is NP-complete for .

## 4 Tractability

In this section we describe a network flow algorithm that solves the possible winner problem in polynomial time for the -approval and rules, when the preference profile is partitioned. Since we assume that the scoring vectors are normalized, then this algorithm is applicable to all -valued scoring rules. Some of the proofs in this section are deferred to the appendix.

##### Maximal Scores

Given a partial order , and a candidate , we denote by the maximum score that candidate can obtain in any linear extension of . That is, . It is straightforward to see that the maximum score of in any extension of is determined by the cardinality of the set of candidates that are preferred to it in . That is,

 smax(o,c)=→α(∣∣c′∈C∣c′≻oc∣∣+1)

where is the scoring vector. We denote by the maximum score that candidate can obtain in any extension of the partial profile to a complete profile. It is straightforward to see that this score can be obtained by maximizing the score for each partial vote independently. Therefore:

 smax(O,c)=n∑i=1smax(oi,c)

When the partial profile is clear from the context then we refer to this score as .

In many cases it is convenient to fix the position of the distinguished candidate in the partial votes such that its score is maximized. Formally, let denote the partial vote. We denote by the partial profile that is consistent with , and where the position of is fixed at the topmost position in each vote. Then:

 oci=oi∪{c≻c′∣c′⊁oic}

In this case, the score of in any extension of is .

##### Elections with Partitioned Preferences

Let be a partitioned partial profile on . Recall that is a partitioned profile if all preferences in are partitioned. Lemma 4 below shows that for deciding whether a distinguished candidate is a possible winner over a profile of partitioned preferences, we may restrict our attention to extensions of the profile where the position (and score) of is fixed to the top of its partition. This is not the case when the profile is not limited to partitioned preferences as shown in the following example. The proof of Lemma 4 is deferred to the appendix.

We consider the election where , , and is the positional scoring rule corresponding to the vector . We consider the problem of deciding whether candidate is a possible winner. The votes are as follows.

 o1 a≻b≻c≻d o2 a≻b≻c≻d o3 b≻a≻c≻d o4 b≻a≻c≻d o5 b≻a

Let denote an extension of in which . For we have that making a possible co-winner. Now consider the extension in which . For we have that , and . Likewise, in the extension in which , is, again, the winner of the election. So we see that despite the fact that is a possible co-winner in , it is not the possible co-winner if positioned at its highest ranking position in every vote.

Let denote an election instance where is a partitioned profile. A distinguished candidate is a possible winner (co-winner) in if and only if it is a possible winner (co-winner) in .

### 4.1 k-approval

Let denote an election where is a partitioned profile, and is the -approval voting rule. As a consequence of Lemma 4, when dealing with partitioned preferences, we may restrict our attention to extensions of where is positioned at the top of its partition in every vote. Specifically, in every profile that extends , candidate gets exactly points. Now, consider any other candidate . If then we have that . Therefore, cannot be a winner (or co-winner) in any complete profile .

Otherwise, if then can top in only if is ranked in positions (i.e., receive points) in at least of the votes in which it could have received a point. Lemma 4.1 below formalizes this condition. The proof is deferred to the appendix.

Let be an election instance where is a partitioned profile, and is the -approval voting rule. Candidate is a possible co-winner in if and only if there exists a complete profile where every candidate is ranked in positions in at least of the votes in which can receive a point (i.e., ).

#### 4.1.1 Network Flow Algorithm

Let be an election where is a partitioned profile, is the -approval voting rule, and is a distinguished candidate. We apply Lemma 4.1 in a maximum network flow algorithm for deciding whether is a possible co-winner in . We begin by describing the network and then prove the correctness of the algorithm.

##### Network Description

The network will contain the following sets of nodes:

1. A source node , and sink node .

2. Candidate nodes : all candidates for which .

3. Vote nodes : For every vote where , the network will contain a single node where () is the partition containing the index . For example, in the vote of Figure 1, the node represents the second partition .

The edges of the network:

1. The set of edges . The capacity of edge is . By construction, the capacity is strictly positive.

2. A candidate node will have outgoing edges to all vote nodes in which it belongs to partition (in which it can lose a point). Formally:

 EVC,VO={(a,oi,j):a∈Aij}

The capacity of every edge in is .

3. The set of edges . The capacity of every edge is set to the number of positions in the partition whose corresponding score is . Formally, . For example, in the vote of Figure 1, the corresponding edge capacity is .

Let be an election where is -approval and is a partitioned profile. A distinguished candidate is a possible co-winner in if and only if the maximum flow in the network is

 ∑{a∈C∣smax(O,a)>smax(O,c)}(smax(O,a)−smax(O,c)) (1)
###### Proof.

The if direction.
Suppose that is a possible winner in . By Lemma 4.1, there exists a complete profile such that every candidate is ranked in positions in at least of the votes in which . That is, in , there exist votes in which while . Since is -approval then in every such vote , candidate is ranked in a position strictly greater than (but smaller than the index corresponding to its partition in ). By the way we constructed the network, there exist at least nodes for which there is a directed edge . Pushing a flow of on these edges, and repeating for every candidate results in the required maximum flow.

The only if direction
So now, assume that we have a maximum network flow (1), and we show how to construct a profile in which is the winner. A maximum flow of (1) implies that every candidate node was able to push all of its incoming flow of to the vote nodes. That is, there exist precisely nodes that received a unit of flow from . In each of the corresponding votes , in which candidate belongs to partition , we position candidate somewhere in the range of positions where it receives a score of . This is possible because given the maximum flow, and according to the capacities assigned to the edges from nodes to , we know that the number of candidates assigned to these positions in the vote does not exceed the capacity of . Repeating this procedure for every candidate node , and placing the rest of the candidates in arbitrary positions, results in a complete ranking that abides to the conditions of Lemma 4.1, making a possible winner in . ∎

Let be an election instance where is the -approval voting rule, , and is a partitioned profile defined as follows.

o1 o2 o3 {b,c,d,e} ≻ {a} {b,c,d} ≻ {a,e} {b,e} ≻ {a,c,d} {b,d} ≻ {a,c,e} {c,d,e} ≻ {a,b} {c} ≻ {a,b,d,e}

The table below presents the number of points each candidate has to lose (with respect to ) so that is the winner.

Candidate smax(O,⋅)−smax(O,c)+1 0 2 2 1

The resulting network is presented in Figure 2. The blue edges can carry a capacity of . Bold edges represent a flow that takes up the capacity of the edge. The flow presented in the figure may correspond to one or more complete profiles in which is the winner.

### 4.2 The Positional Scoring rule (2,1,1,…,1,0)

We now consider an election where is the rule and is a partitioned profile. As usual, is our distinguished candidate. It has been shown that, in general, the Possible-Winner problem for is NP-complete [2]. We show that the network flow algorithm of the previous section solves this problem in polynomial time if is a partitioned profile.

Let denote a partitioned vote and a candidate with a maximum score in . If has two or more partitions then in any extension exactly one of the following can occur: (1) , (2) and , (3) or (4) and . In all of these options, candidate can lose either 0 or 1 points in . Formally, for any candidate , and any partitioned vote with at least two partitions we have that .

Now, let us assume that is a partitioned preference with a single partition. That is, contains no precedence constraints. In this case, by Lemma 4, we can assume that any complete profile in which wins (or co-wins), is an extension of . In particular, this means that we may assume that in , candidate is ranked in the topmost position and thus receives two points (i.e., ). This, in turn, means that for any other candidate exactly one of the following can occur: (1) (2) and . As in the previous case, candidate can lose either 0 or 1 points in . Formally, .

Now that we have established that every candidate can “lose” at most one point in every vote, we can apply the network flow algorithm of the previous section.

## 5 Hardness

Let be a pure, positional scoring rule. From this point on we assume that produces a scoring vector with at least distinct values (the case of -valued scoring rules was considered in the previous section).

([7]) We say that a voting rule is differentiating if there exists some constant such that for all the score vector contains two positions where such that . Dey and Misra [7] have shown that the possible winner problem is NP-complete for all differentiating scoring rules. The proof (Theorem 6) relies only on partitioned preferences, implying hardness of the Possible-Winner problem for differentiating scoring rules with partitioned profiles. Therefore, we restrict our attention to non-differentiating scoring rules. Formally, for every scoring vector , and for every pair of consecutive values , we have that .

A common strategy in proving hardness for the PW problem is to construct a profile , consisting of a set of linear orders, that enables determining the score of every candidate in according to the requirements dictated by the reductions [3, 7, 1, 2]. Once such a set is constructed, the profile is enhanced with a set of partial votes , where the maximum scores of the candidates are restricted according to the linear votes in . Lemma 5 below [7] states that such a profile can be constructed in polynomial time.

[[7]] Let , be a set of candidates, and a scoring vector of length . Then for every integer vector , there exists a and a voting profile such that for all , and for all . Moreover, the number of votes in is polynomial in .

Our NP-hardness proofs rely on reductions from the NP-complete 3-Dimensional-Matching problem (3DM[8]. The 3DM problem is defined as follows. We are given three disjoint sets , , and each containing exactly elements, and a set of triples. We wish to know whether there is a subset of disjoint triples that covers all elements of .

In some of our theorems, we will need functions that map each instance of 3DM to a natural number, and in some sense behave like a polynomial. For this sake, we call

 f:{I∣I is an instance of \textsc3DM}↦N

a poly-type function for 3DM [3] if the function value is bounded by a polynomial in for every input instance of 3DM.

A 3DM instance can be reduced to a Possible-Winner instance for a scoring rule which produces a size- scoring vector that fulfills the following. There is an such that with , and . A suitable poly-type function for 3DM can be computed in polynomial time.

###### Proof.

Let denote the value that occupies positions in . By the previous discussion, and since is non-differentiating, the scoring vector contains three indexes , , and such that , , . Schematically:

 (2)

Let denote a 3DM instance where . The set of candidates is defined by where denotes the distinguished candidate, the set of candidates that represent the elements of the 3DM instance, and and contain disjoint candidates such that the following hold. We define where the sets are pairwise disjoint, and for all . The sets

will be used for “padding” some positions relevant to the construction. The set

contains candidates needed to pad irrelevant positions. We set . Recall that . Intuitively, this means that the portion of the scoring vector , occupied by values different from , is large enough to contain all elements besides one of the sets .

For every triple let such that (see (2)). We construct the following linear vote .

 vs=−−−−−→(C∖Cs)≻x≻y≻−−−→(Hs)≻z≻−−→(Cs)

where , , and are arbitrary complete orders over the candidate sets , , and respectively. Using we define the partial partitioned vote as follows.

 v′s=−−−−−→(C∖Cs)≻(x∪y∪Hs∪z)≻−−→(Cs)

Note that this implies that in any extension of , items will occupy the positions in the range .

We denote by and . By Lemma 5 there exists a set of linear votes , of size polynomial in , where the scores of the candidates in the combined profile are as follows:

 sP∪Q(x) =sP∪Q(c)+2 ∀x∈X sP∪Q(y) =sP∪Q(c)−1 ∀y∈Y sP∪Q(z) =sP∪Q(c)−1 ∀z∈Z sP∪Q(h) =sP∪Q(c) ∀h∈H sP∪Q(d)

We observe that the score of is the same in any extension of and is identical to its score in . We define the instance of Possible-Winner to be , and proceed with the reduction.

In the forward direction, suppose that is a instance of 3DM. Then, there exists a collection of disjoint sets in such that . For every we extend the partial vote to as follows.

 ¯¯¯¯¯v′s=⎧⎨⎩−−−−−→(C∖Cs)≻y≻−−−→(Hs)≻z≻x≻−−→(Cs)s∈S′−−−−−→(C∖Cs)≻x≻y≻−−−→(Hs)≻z≻−−→(Cs)s∉S′

where, again, is an arbitrary complete order over the candidates . We consider the extension of to . We claim that is a co-winner in the profile because:

1. For all :

2. For all :

3. For all : .

4. For all : .

5. For all : .

For the reverse direction, suppose that the Possible-Winner instance is a instance. Then there exists an extension of the set of partial, partitioned votes to a set of complete votes such that is a co-winner in . We refer to the extension of as .

We recall that the score of is the same in any extension of and is identical to its fixed score in . We first claim that for every in which is not in position (see (2)), then occupies this position. Indeed, if occupied this position then its score would increase by at least (i.e., compared to the vote ). Since cannot lose points in any of the votes this would mean that . But then we arrive at a contradiction that is a co-winner. Likewise, if some occupied this position then its score would increase by at least because, according to the construction, the position of is fixed in the rest of the votes, and thus, it cannot lose points in any other vote in . Then, contradicting the assumption that is a co-winner. Finally, candidates in cannot occupy position in any extension of .

We now claim that, for every , there exists exactly one triple such that is not in position in . Otherwise, by the previous claim, there must be at least two votes in which position is occupied by candidates from . Since our profile contains more than votes (i.e., )), then by the pigeon-hole principle there exists a candidate that appears in position at least twice. But then, the overall score of candidate must have increased by strictly more than point. However, in such a scenario, the score of will be strictly more than the score of contradicting the fact that is a co-winner in .

Now, the claim follows from the observation that every must lose points in order for to co-win. From the claim that there is exactly one vote in which does not occupy position for every , and since , we have that contains precisely votes corresponding to triples in which does not occupy position . Furthermore, for every it must be the case that loses two points and thus .

We now show that . It is clear that , so we show that . Assume the contrary. Then there is a candidate that does not belong to . If , then by the claim that in every vote in which is not in position , there is some candidate in this position, and that , there must be some candidate that appears at least twice in this position. But then , and we arrive at a contradiction that is a co-winner. If then by the same reasoning there is some candidate that appears at least twice in position (see (2)). But then , and we arrive at a contradiction that is a co-winner. ∎

A 3DM instance can be reduced to a Possible-Winner instance for a scoring rule which produces a size- scoring vector that fulfills the following. There is an