DeepAI

# The Complexity of Determining the Necessary and Possible Top-k Winners in Partial Voting Profiles

When voter preferences are known in an incomplete (partial) manner, winner determination is commonly treated as the identification of the necessary and possible winners; these are the candidates who win in all completions or at least one completion, respectively, of the partial voting profile. In the case of a positional scoring rule, the winners are the candidates who receive the maximal total score from the voters. Yet, the outcome of an election might go beyond the absolute winners to the top-k winners, as in the case of committee selection, primaries of political parties, and ranking in recruiting. We investigate the computational complexity of determining the necessary and possible top-k winners over partial voting profiles. Our results apply to general classes of positional scoring rules and focus on the cases where k is given as part of the input and where k is fixed.

• 4 publications
• 34 publications
05/18/2020

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## 1 Introduction

A central task in social choice is that of winner determination—how to aggregate the candidate preferences of voters to select the winner. Relevant scenarios may be political elections, document rankings in search engines, hiring dynamics in the job market, decision making in multiagent systems, determination of outcomes in sports tournaments, and so on [5]. Different voting rules can be adopted for this task. The computational social-choice community has studied in depth the family of the positional scoring rules, where each voter assigns to each candidate a score based on the candidate’s position in the voter’s ranking, and then a winner is a candidate who receives the maximal sum of scores. Famous instantiations of the positional scoring rules include the plurality rule (where a winner is most frequently ranked first), the veto rule (where a winner is least frequently ranked last), their generalizations to -approval and -veto, respectively, and the Borda rule (where the score is the actual position in the reverse order).

The seminal work of Konczak and Lang Konczak2005VotingPW has addressed the situation where voter preferences are expressed or known in just a partial manner. The framework is based on the notions of the necessary winners and possible winners, who are the candidates that win in every completion, or at least one completion, respectively, of the given partial preferences into complete ones. More precisely, a voting profile consists of a partial order for each voter, and a completion consists of a linear extension for each of the partial orders. Determining the necessary and possible winners is computationally challenging since, conceptually, it involves reasoning about the entire (exponential-size) space of such completions. The complexity of these problems has been thoroughly studied in a series of publications that established a full classification of a general class of positional scoring rules (the “pure” scoring rules) into tractable and intractable ones [4, 21, 3].

The outcome of an election often goes beyond the single winner to the set of top- winners. For example, the top- winners might be the elected parliament members, the entries of the first page of the search engine, the job candidates to recruit, and the finalists of a sports competition. In the case of a positional scoring rule, the top- winners are the candidates who receive the top scores (under some tie-breaking mechanism) [16]. Adopting the framework of Konczak and Lang Konczak2005VotingPW, in this paper we investigate the computational complexity of determining the necessary and possible top- winners for incomplete voting profiles and positional scoring rules.

We show that the top- variant makes the problems fundamentally harder than their top- counterparts (necessary and possible winners) when is given as input. For example, it is known that detecting the possible winners is NP-hard for every pure rule, with the exception of plurality and veto where the problem is solvable in polynomial time [4, 21, 3]; we show that in the case of top-, the problem is NP-hard for every pure rule, including plurality and veto. Moreover, tractability of the necessary winners does not extend to the necessary top- winners: we show that the detecting whether a candidate is necessarily a top- winner is coNP-complete for a quite general class of positional scoring rules that include all of the aforementioned ones. We also study the impact of fixing and establish a more positive picture: detecting the necessary top- winners is tractable (assuming that the scores are polynomial in the number of candidates) and detecting the possible the top- winners is tractable for plurality and veto.

The concept of the top- winners can be viewed as a special case of multiwinner election that has been studied mostly in the context of committee selection. Various utilities have been studied for qualifying selected committee, such as maximizing the number of voters with approved candidates [1] and, in that spirit, the Condorcet committees [10, 8], aiming at proportional representation via frameworks such as Chamberlin and Courant’s 10.2307/1957270 and Monroe’s monroe_1995, and the satisfaction of fairness and diversity constraints [7, 6].

In the case of incomplete voter preferences, the generalization of the problem we study is that of detecting the necessary and possible committee members. These are interesting and challenging problems in all the variants of committee selection, and we leave them for future investigation. Note, however, that the problem of determining the elected committee can be intractable even if the preferences are complete [17, 18, 8, 20], in contrast to the top- winners. Yet, we show that our results imply the tractability of determining whether a candidate set is a necessary or possible Condorcet committee in the case of the plurality and veto rules. The problem of multiwinner determination for incomplete votes has been studied by Lu and Boutilier DBLP:conf/ijcai/LuB13 in a perspective different from the necessary and possible top- winners: find a committee that minimizes the maximum objection (or “regret”) over all possible completions.

## 2 Preliminaries

We begin with some notation and terminology.

##### Voting Profiles and Positional Scoring Rules.

Let be the set of candidates (or alternatives) and let be the set of voters. A voting profile consists of linear orders on , where each represents the ranking of by .

A positional scoring rule is a series of

-dimensional vectors

where and . We denote by . Some examples of positional scoring rules include the plurality rule , the -approval rule that begins with ones, the veto rule , the -veto rule that ends with zeros, and the Borda rule .

Given a voting profile , the score that the voter contributes to the candidate is where is the position of in . The score of in is or simply if is clear from context. The winners (or co-winners) are the candidates with a maximal .

We make standard assumptions about the positional scoring rule . We assume that is computable in polynomial time in . We also assume that the numbers in each are co-prime (i.e., their greatest common divisor is one).

A positional scoring rule is pure if is obtained from by inserting a score at some position, for all .

##### Partial Profiles.

A partial voting profile consists of partial orders on set of candidates, where each represents the incomplete preference of the voter . A completion of is a complete voting profile where each is a completion (i.e., linear extension) of the partial order .

The problems of necessary winners and possible winners were introduced by Konczak and Lang Konczak2005VotingPW. Given a partial voting profile , a candidate is a necessary winner if is a winner in every completion of , and is a possible winner if there exists a completion of where is a winner. The decision problems associated to a positional scoring rule are those of determining, given a partial profile and a candidate , whether is a necessary winner and whether is a possible winner. We denote these problems by and , respectively. A classification of the complexity of these problems has been established in a sequence of publications.

###### Theorem 1 (Classification Theorem [4, 21, 3]).

can be solved in polynomial time for every positional scoring rule. is solvable in polynomial time for plurality and veto; for all other pure scoring rules, it is NP-complete.

In this paper, we aim towards generalizing the Classification Theorem to determine the necessary and possible top- winners, as we formalize next.

##### Top-k Winners.

In principle, a top- winner is a candidate that is ranked at one of the top places with respect to the sum of scores from the voters. However, for a precise definition, we need to reason about ties. One could adopt several options for being on the top- winners: (a) w.r.t. at least one tie-breaking order; (b) w.r.t. every tie-breaking order; and (c) w.r.t. a tie-breaking order given as input. For simplicity of presentation, we adopt the third variation and assume that the tie-breaking order is given as input. Nevertheless, all of our results hold for all three variations.

Formally, let be a positional scoring rule, be a set of candidates, a voting profile, and a tie breaker, which is simply a linear order over . Let be the linear order on that sorts the candidates lexicographically by their scores and then by ; that is,

 RT:= {c1>c2:s(T,c1)>s(T,c2)}∪ {c1>c2:s(T,c1)=s(T,c2)∧c1≻tiec2}.

A candidate is a top- winner if the position of in , denoted by , is at most . Note that a top-1 winner is necessarily a winner, but a winner might not be a top-1 winner due to tie breaking.

If is replaced with a partial voting profile , then a candidate is a necessary top- winner if is a top- winner in every completion of , and a possible top- winner if is a top- winner in at least one completion of . Hence, for a positional scoring rule , we have two computational problems where the input consists of a candidate set , a partial profile , a tie breaker , a candidate and a number :

• In , the goal is to determine whether is a necessary top- winner.

• In , the goal is to determine whether is a possible top- winner.

We will also consider the versions where is fixed, and then denote it by parameterizing the problem with : and

We use the following notation. For a set and a partition of :

• denotes the partitioned partial order .

• denotes an arbitrary linear order on that completes .

A linear order is also denoted as a vector . The concatenation is .

## 3 Hardness of Top-k Winners

We first show that the problems we study are computationally hard for quite general classes of positional scoring rules.

### 3.1 Plurality and Veto

The following theorems state the hardness for the plurality and veto rules where both and are solvable in polynomial time (according to the Classification Theorem).

###### Theorem 2.

For the plurality rule, is coNP-complete and is NP-complete.

###### Proof.

Memberships in the corresponding classes (coNP and NP) are straightforward, so we prove only hardness. We show a reduction for each of the two problems.

:  We show a reduction from exact cover by-3sets (X3C), which is the following decision problem: Given a vertex set and a collection of 3-element subsets of , can we cover all the elements of using pairwise-disjoint sets from ? For , denote by the set of edges incident to .

Given and , we construct an instance of under the plurality rule where , where , and where is the partial voting profile . For every ,

 Pi=P(E(u),E∖E(u),{c∗}).

This means that the th voter can vote only for edges that cover . For the order is . To complete, we show that there is an exact cover if and only if is not a necessary top- winner.

Suppose that is not a necessary top- winner, that is, there are candidates and a completion of such that for all . Since every edge can get at most three votes, we get that for every . Therefore is an exact cover: every vertex is covered by the edge that voted for, and the edges are pairwise disjoint (since, if two edges are overlapping, then one gets at most two votes).

Conversely, given an X3C solution define a profile such that for every we have

 Ti=O(Q∩E(ui),E(ui)∖Q,E∖E(ui),{c∗}).

Every extends , for all we have , and . Hence, all edges in defeat , and is a not a top- winner in .

:  We use a reduction from the dominating set problem, which is the following: Given an undirected graph and an integer , is there a set of size such that every vertex is either in or adjacent to some vertex in ? Given a graph with , we construct an instance for under plurality where , where , and where . Let be the set of neighbours of , and let . For all we have

Hence, the th voter can vote only for vertices that dominate . To complete, we show that the graph has a dominating set of size if and only if is a possible top- winner.

Suppose there is a dominating set of size , consider the profile where for every ,

 Ti:=O(N(ui)∗∩D,N(ui)∗∖D,U∖N(ui)∗,{c∗}).

In this completion, for each we get . These are candidates that defeats, therefore is a possible top- winner. Conversely, if is a possible top- winner then in some completion it defeats at least candidates, and these candidates have a score 0 in . Let be the set of candidates that does not defeat in , all voters voted for candidates in and . A voter can only vote for vertices which dominate , hence is a dominating set of size at most . ∎

Next, we show the hardness of and beyond the plurality rule. Given a binary positional scoring rule , we define the complementary-reversed scoring rule, denoted , to be the one given by . For example, the complementary-reversed rule of plurality is veto, and more generally, the complementary-reversed rule of -approval is -veto.

###### Lemma 1.

For every binary positional scoring rule , there is a reduction

1. from for to the complement of for ;

2. from for to the complement of for .

###### Proof.

For a partial order , the reversed order is defined by . Note that extends if and only if extends .

Given as input under with , consider under where . Let be a completion of , observe the completion of . For every candidate and a voter we get so overall . Since the tie-breaking order is also reversed, it holds that for every . In same way, if is a completion of then by reversing the orders we get a completion of such that for every . We can deduce that for any candidate and integer ,

1. is not a necessary top- winner w.r.t and (there exists a completion of such that ) if and only if is a possible top- winner w.r.t and (there exists a completion of such that ).

2. is a possible top- winner w.r.t and (there exists a completion of such that ) if and only if is not a necessary top- winner w.r.t and (there exists a completion of such that ).

From the above two points we conclude the two parts of the lemma, respectively. ∎

Combining Lemma 1 and Theorem 2, we conclude that:

###### Theorem 3.

For the veto rule, is coNP-complete and is NP-complete.

### 3.2 Beyond Plurality and Veto

What about positional scoring rules other than plurality and veto? For any other pure positional scoring rule, is NP-complete by the Classification Theorem, so is also NP-complete (by choosing ). Combining this observation with Theorems 2 and 3, we conclude that:

###### Corollary 1.

is NP-complete for every pure positional scoring rule.

While we do not have a full classification for , we show the hardness of under general conditions that include the commonly studied rules. First, we can deduce hardness for every pure positional scoring rule with binary scores. We already established hardness for plurality and veto in Theorems 2 and 3. For any other rule in this class, is NP-complete for (by the Classification Theorem), so a small change in the proof of Lemma 1 shows that is coNP-complete for . We conclude that:

###### Corollary 2.

is coNP-complete for every pure positional scoring rules with binary scores.

To discuss rules with scores beyond binary, we define the class of polynomially frequent scoring rules where some score occurs frequently in the scoring vector. All commonly studied rules fall under this definition, except for Borda.

###### Definition 1.

A positional scoring rule is polynomially frequent if there exists a score and a constant such that .

Examples of polynomially frequent rules include -approval (where ) and -veto (where ). Another rule is that has been studied in depth [2]. This class strictly generalizes that of the almost constant scoring rules that has also been studied in the context of the complexity of winner determination [14, 13].

We will prove that is hard for all pure polynomially frequent scoring rules. For that, we need a definition and a lemma. Let and be two positional scoring rules. We say that polynomially contains if there exist a polynomial , an index and two numbers and such that for all . For instance, -approval polynomially contains plurality for every fixed , by choosing , , and . Similarly, -veto polynomially contains veto.

###### Lemma 2.

Let and be two positional scoring rules. If polynomially contains , then there is a reduction

1. from for to for ;

2. from for to for .

###### Proof.

Let , , , and be the functions that realize the polynomial containment. Given as input under with , consider the input under where:

• where and are disjoint sets of new candidates with and . Note that .

• where, in each , the highest-ranked candidates are the ones of , the lowest-ranked candidates are the ones of , and between and the candidates of are the same as in . In our notation, .

• In , the highest-ranked candidates are the ones of , the lowest-ranked candidates are the ones of , and between and the candidates of are the same as in . In our notation, .

Let be a completion of . Observe the completion of where . Since polynomially contains in , for every we get that

 s(T′,c,r′) =n∑j=1s(T′j,c,r′)=n∑j=1am⋅s(Tj,c,r)+bm =am⋅s(T,c,r)+nbm.

Hence, by the definition of , the positions of the candidates satisfy . Conversely, let be a completion of . Every has to be of the form , so removing and from all the linear orders gives a completion of such that for all . We conclude that for all and it holds that is a necessary (resp., possible) top- winner w.r.t.  and if and only if is a necessary (resp., possible) top- winner w.r.t.  and . ∎

Note that Lemma 2 can be applied for non-pure rules, so the second item can be used for rules not covered by Corollary 1. Consequently, we get the following general hardness.

###### Theorem 4.

is coNP-complete for every pure polynomially frequent scoring rule.

###### Proof.

Let be a pure polynomially frequent scoring rule, and denote by . Let be the minimal where occurs in , and let be the index of in . First, consider the case where , which means that contains both and a score greater than . Since the rule is pure we can deduce that for every , also contains both and a score greater than . Then, there exists some score such that contains the vector of length . Choosing , as the biggest index of , and shows that, in this case, polynomially contains plurality. The hardness results then follow from Theorem 2 and Lemma 2. Now consider the case where , hence for every , contains both and a score smaller than (since the rule is pure). Then, there exist some score such that contains the vector of length . Choosing , as the smallest index of , shows that in this case polynomially contains veto. Hardness for then follows from Theorem 3 and Lemma 2. ∎

To complete the picture, we are still missing the complexity of the top- winners for the (pure) positional scoring rules that are not polynomially frequent. An example that stands out is the Borda rule. This is left as an open direction for future investigation that we have found quite challenging. For the special case of Borda, we can prove the hardness of .

###### Theorem 5.

is coNP-complete for the Borda rule.

The proof, discussed next, is nontrivial and heavily relies on the specific structure of this rule.

#### 3.2.1 Proof of Theorem 5

We use the technique of circular voting blocks of Baumeister, Roos and Jörg DBLP:conf/atal/BaumeisterRR11. For a set and , the th circular vote is

 Mi(A):=(ai,ai+1,…,at,a1,a2,…,ai−1).

We reduce from X3C as defined in the proof of Theorem 2. Given an X3C instance with and , we construct an input for under Borda. The candidates are where and . The voting profile is the concatenation (union) of the three parts described next.

First, . For every , only edges that cover can receive a score of . Edges that do not cover can receive at most , and receives . Formally, denote by the degree of in the graph and let and . We assume that , otherwise the problem is trivial. The partial order is . Note that for every , the number of candidates ranked above in is , so indeed it receives a score of at most .

Second, is composed of copies of the profile . For every the order is constructed in the following way. Start with , then insert and such that the score of is and scores of are accordingly if is even. If

is odd then the scores of the edges are the same as before but with

instead of and 1 instead of 0. Note that in both cases, the sum of the scores of the edges is the same.

Finally, consists of copies of the profile . For this part only, denote . For every , .

We state some observations regarding the profile. In , the score of is

 s(T2,c∗)=Scvr⋅s((T21,…,T2m),c∗)=Scvr⋅m2

For every , the score in is

 s(T2,e) =Scvrm∑i=2(2i−1)=Scvr(m2−1) =s(T2,c∗)−Scvr.

In , for every , the score is

 s(T3,d) ≤4(3q+Scvr)(m+1)(m−2)≤4m2(3q+Scvr)

For every , the score in is

 s(T3,c)=4(3q+Scvr)m+1∑i=1(2m−i) =2(3q+Scvr)(3m2+m−2)≥6m2(3q+Scvr).

Hence, for every pair it holds that . Let be a completion of . For every pair , the score in satisfy and (by the definition of the Borda rule). Combining these two inequalities with what we showed for , we can deduce that , which means that the candidates in always defeat all candidates in .

###### Claim 1.

If there is an X3C, then is not a necessary top- winner.

###### Proof.

Assume, w.l.o.g., that the exact cover is and every edge in the cover is . Define a completion , . In , every edge in the cover receives the following scores:

• gets from , that is, is placed at the top in the vertices which it covers.

• gets from the vertices of , gets from the vertices of and so on (when we reach we go to and continue until ). Note that in this way there is no vertex that should give the same score to two different edges, and since the range of scores gets is as required by the definition of .

For every edge , the total score in is and recall that . Combining this with what we already know for and implies that

 s(T,e)−s(T,c∗) =Scvr−Scvr+0=0.

All candidates in defeat , hence is not a necessary top- winner. ∎

###### Claim 2.

If is not a necessary top--winner, then there is an X3C.

###### Proof.

Let be a completion where at least candidates defeat , let be the highest rated candidates in . The candidate always defeats all candidates in , hence . We show a lower bound and an upper bound on the total score of in .

Lower bound.  As we already showed, every should get in order to defeat , therefore

 ∑e∈Qs(T1,e) ≥q⋅Scvr=3q⋅(2m−1+q−1∑i=1(m−i)) =3mq2+3mq−3q32+3q22−3q

Upper bound.  For every denote by the degree of in the sub-graph induced by , then . We get that

 ∑e∈Qs(T1,e) ≤3q∑i=1⎛⎝degQ(ui)∑j=1(2m−j)+q−degQ(ui)∑j=1(m−j)⎞⎠ = 3mq2+3mq−3q32+3q22−3q∑i=1degQ(ui)2.

Overall, the two bounds imply that and this is possible only if all the degrees are one: When all degrees are one, the sum is exactly . If we decrease to zero and increase to two, then the total sum increases by three. Any further changes cannot decrease the total sum. Therefore, all degrees in the sub-graphs induced by are one, and is an X3C. ∎

### 3.3 Top-k Sets

We have shown hardness results for the necessary and possible top- winners. Interestingly, we can retain the tractable cases of the necessary-winner and possible-winner problems for the variant of the problem where we are given a set of candidates, and the goal is to determine whether constitutes the exact set of top- winners. We say that is a necessary top- set if is the set of top- winners in every completion, and a possible top- set if is the set of top- winners in at least one completion.

###### Theorem 6.

Let be a positional scoring rule. We can determine in time :

1. whether a given candidate set is a necessary top- set;

2. whether a given candidate set is a possible top- set, assuming that is either plurality or veto.

###### Proof.

For the first part of Theorem 6 (necessity), we only need to determine whether a candidate outside of can defeat a candidate from . This can be done using the algorithm of Xia and Conitzer DBLP:journals/jair/XiaC11, with a minor adjustment to account for tie breaking.

For the second part (possibility), for every candidate and every integer score we use Lemma 5 (that we prove in the following section) to check if there exists a completion which satisfies the following conditions. First, . Second, for every , if then , otherwise . This means that all candidates in defeat . Finally, For every , if then , otherwise . This means that defeats all candidates in . is a possible top- set if and only if such completion exists for some candidate and score . ∎

Interestingly, for the plurality and veto rules, a set of candidates is a top- set for at least one tie-breaking order if and only if is a Condorcet committee [11]. Then, by a simple adjustment of the proof of Theorem 6 we can conclude that, in the case of plurality and veto, one can determine in polynomial time whether a given candidate set is a necessary or possible Condorcet committee.

## 4 The Case of a Fixed k

In the previous section, we established that the problems of finding the necessary and possible top- winners are very often intractable. In this section, we investigate the complexity of these problems under the assumption that is fixed (and, in particular, can be the degree of the polynomial that bounds the running time). We will show that the complexity picture for and is way more positive, as we generalize the tractability of almost all of the tractable scoring rules for and . We will also generalize hardness results from to ; interestingly, this generalization turns out to be quite nontrivial.

### 4.1 Tractabiliy of NTW⟨k⟩

We first prove that is tractable for every positional scoring rule (pure or not), as long as the scores are bounded by a polynomial in the number of candidates; in this case, we say that the rule has polynomial scores. Note that this assumption is in addition to our usual assumption that the scores can be computed in polynomial time.

###### Theorem 7.

For all fixed and positional scoring rules with polynomial scores, is in polynomial time.

Note that all of the specific rules mentioned so far (i.e., -approval, -veto, Borda and so on) have polynomial scores, and hence, are covered by Theorem 7. An example of a rule that is not covered is the rule defined by .

In the remainder of this section, we prove Theorem 7. To determine whether a candidate is a necessary top- winner, we search for a counterexample, that is, candidates that defeat in some completion. For that, we iterate over every subset and determine whether these candidates can get a combination of scores that constitues the counterexample.

More formally, let be a set of candidates and a positional scoring rule. For a partial profile and a sequence of candidates from , we denote by the set of all possible scores that the candidates in can obtain jointly in a completion:

 π(P,S):={(s(T,c1),…,s(T,cq)):T completes P}

Note that . When consists of a single voter , we write instead of .

A counterexample for being a necessary top- winner is a sequence where and , and a sequence