# On the Complexity of Constructive Control under Nearly Single-Peaked Preferences

We investigate the complexity of Constructive Control by Adding/Deleting Votes (CCAV/CCDV) for r-approval, Condorcet, Maximin and Copeland^α in k-axes and k-candidates partition single-peaked elections. In general, we prove that CCAV and CCDV for most of the voting correspondences mentioned above are NP-hard even when k is a very small constant. Exceptions are CCAV and CCDV for Condorcet and CCAV for r-approval in k-axes single-peaked elections, which we show to be fixed-parameter tractable with respect to k. In addition, we give a polynomial-time algorithm for recognizing 2-axes elections, resolving an open problem. Our work leads to a number of dichotomy results. To establish an NP-hardness result, we also study a property of 3-regular bipartite graphs which may be of independent interest. In particular, we prove that for every 3-regular bipartite graph, there are two linear orders of its vertices such that the two endpoints of every edge are consecutive in at least one of the two orders.

## Authors

• 6 publications
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We consider the problem of partitioning a graph into a non-fixed number ...
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We prove a complexity dichotomy theorem for a class of Holant problems o...
11/18/2020 ∙ by Austen Z. Fan, et al. ∙ 0

• ### Recognizing Single-Peaked Preferences on an Arbitrary Graph: Complexity and Algorithms

This paper is devoted to a study of single-peakedness on arbitrary graph...
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• ### Election Control by Manipulating Issue Significance

Integrity of elections is vital to democratic systems, but it is frequen...
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• ### Relation-algebraic and Tool-supported Control of Condorcet Voting

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## 1 Introduction

Constructive Control by Adding Votes (CCAV) and Constructive Control by Deleting Votes (CCDV) are two of the election control problems studied in the pioneering paper by Bartholdi III, Tovey, and Trick [Bartholdi III et al. (1992)]. These two problems model the applications where an election controller aims to make a distinguished candidate a winner by adding or deleting a limited number of voters. Since their seminal work, the complexity of these problems under a lot of prestigious voting correspondences have been studied, and it turned out that many problems are NP-hard [Bartholdi III et al. (1992), Faliszewski et al. (2011a), Faliszewski et al. (2009), Lin (2012)]. However, when restricted to single-peaked elections, many of them become polynomial-time solvable [Brandt et al. (2015), Faliszewski et al. (2011b)]. Recall that an election is single-peaked if there is an order of the candidates, the so-called axis, such that each voter’s preference purely increases, or decreases, or first increases and then decreases along this order. A natural question is that, as preferences of voters are extended from the single-peaked domain to the general domain with respect to a certain concept of nearly single-peakedness, where does the complexity of these problems change? A large body of results have been reported with respect to some nearly single-peaked domains. This paper aims to extend these study by investigating the complexity of the above two problems under several important voting correspondences when restricted to the -axes and -candidates partition single-peaked elections (-axes and -CP elections for short respectively). Generally speaking, an election is a -axes election if there are  axes such that every vote is single-peaked with respect to at least one of the axes. Equivalently, an election is a -axes election if the votes can be partitioned into  sets each of which induces a single-peaked election. An election is a -CP election if there is a -partition of the candidates such that the subelection restricted to each  is single-peaked. Clearly, -axis elections and -CP elections are exactly single-peaked elections. The voting correspondences studied in this paper include -approval, Condorcet, Maximin, and Copeland, where is a rational number such that .

Additionally, we also resolve an open question regarding the complexity of recognizing -axes elections by deriving a polynomial-time algorithm.

### 1.1 Related Work

Our study is clearly related to [Brandt et al. (2015), Faliszewski et al. (2011b)] where many voting problems including particularly CCAV and CCDV for -approval, Condorcet, Maximin, and Copeland were shown to be polynomial-time solvable when restricted to single-peaked elections. Resolving an open question, Yang [Yang (2017b)] recently proved that CCAV and CCDV for Borda are NP-hard even when restricted to single-peaked elections.

Our study is also closely related to the work of Yang and Guo [Yang and Guo (2014a), Yang and Guo (2017), Yang and Guo (2018)] where CCAV and CCDV for numerous voting correspondences in elections of single-peaked width at most  and -peaked elections were studied. Generally, an election has single-peaked width  if the candidates can be divided into groups, each of size at most , such that every vote ranks all candidates in each group consecutively and, moreover, considering each group as a single candidate results in a single-peaked election. An election is -peaked if there is an axis  such that for every vote  there is a -partition of  such that  restricted to each component of the partition is single-peaked. Obviously, -CP elections are a subclass of -peaked elections. In addition, it is known that any election of single-peaked width  is a -CP election for some  [Erdélyi et al. (2017)]. However, there are no general relation between -axes elections and -CP elections, and between -axes elections and elections with single-peaked width  [Erdélyi et al. (2017)].

In addition to CCAV and CCDV, many other problems restricted to single-peaked or nearly single-peaked domains have been extensively and intensively studied in the literature in the last decade (see, e.g., [Betzler et al. (2013), Cornaz et al. (2012), Cornaz et al. (2013), Yu et al. (2013)] for Winner Determination, [Walsh (2007)] for Possible/Necessary Winner Determination, [Yang (2015b)] for Manipulation, [Menon and Larson (2016)] for Bribery, and [Faliszewski et al. (2014), Yang (2017a)] for some other important strategic voting problems). Approval-Based multiwinner voting problems restricted to analogs of single-peaked domains have also been investigated from the complexity perspective very recently [Elkind and Lackner (2015), Liu and Guo (2016), Peters (2018)].

Finally, we point out that a parallel line of research on the complexity of single-crossing and nearly single-crossing domains has advanced immensely too (see, e.g., [Magiera and Faliszewski (2017), Skowron et al. (2015)]).

We also refer to the book chapters [Elkind et al. (2017), Hemaspaandra et al. (2016)] and references therein for important development on these studies.

### 1.2 Our Contributions

Our contributions are summarized as follows.

• We study CCAV and CCDV in -axes and -CP elections under -approval, Condorcet, Copeland, and Maximin.

• We show that many problems already become NP-hard even when  is a very small constant. However, there are several exceptions. (See Table 1 below for the concrete results.) In addition, our results reveal that from the parameterized complexity point of view, CCAV and CCDV for some voting correspondences behave completely differently. For instance, for -approval, CCAV in -axes elections is fixed-parameter tractable (FPT) with respect to , but CCDV is already NP-hard even for , meaning that CCDV restricted to -axes elections is even para-NP-hard with respect to . Our results also reveal that when restricted to different domains, the same problem may behave differently. For instance, for Condorcet, we show that both CCAV and CCDV in -axes elections are FPT with respect to , but they become para-NP-hard with respect to  when restricted to -CP elections. Finally, we would like to point out that our study also leads to numerous dichotomy results for CCAV and CCDV with respect to the values of .

• We study the complexity of determining whether an election is a -axes election. It is known that for , the problem is polynomial-time solvable [Bartholdi III and Trick (1986), Doignon and Falmagne (1994), Escoffier et al. (2008)]. Erdélyi, Lackner, and Pfandler [Erdélyi et al. (2017)] proved that the problem is NP-hard for every . We complement these results by showing that determining whether an election is a -axes election is polynomial-time solvable, filling the last complexity gap of the problem with respect to .

## 2 Preliminaries

In this section, we give the notions used in the paper. For a positive integer , we use  to denote the set of all positive integers no greater than .

Election. An election is a tuple , where  is a set of candidates and  a multiset of votes, defined as permutations (linear orders) over . For two candidates and a vote , we say  is ranked above  or  prefers  to  if . For two subsets of candidates, a vote with preference means that this vote prefers every to every . For brevity, we use for . Here,  is the position of  in , i.e., . For and a vote , let . In addition, let  be  restricted to  so that for , implies . Let . Hence, is the election restricted to . We use to denote the number of votes preferring  to  in . We drop  from the notation when it is clear from the context which election is considered. For two candidates  and  in , we say  beats  if , and  ties  if .

For a linear order  over a set , we say that two elements in  are consecutive if there are no other elements from  between them in the order.

Voting correspondence. A voting correspondence  is a function that maps an election to a non-empty subset  of . We call the elements in  the winners of  with respect to . The following voting correspondences are related to our study.

-approval

Each vote approves exactly its top- candidates. Winners are those with the most total approvals. Throughout this paper,  is assumed to be a constant, unless stated otherwise.

Borda

Every vote  gives points to every candidate  and the winners are the ones with the highest total score. Here,  is the number of candidates.

Copeland ()

For a candidate , let  (resp. ) be the set of candidates beaten by  (resp. tie with ). The Copeland score of  is . A Copeland winner is a candidate with the highest score.

Maximin

The Maximin score of a candidate  is . Maximin winners are those with the highest Maximin score.

Condorcet

The Condorcet winner of an election is the candidate that beats all other candidates. It is well-known that each election has either zero or exactly one Condorcet winner. In addition, both Copeland and Maximin select only the Condorcet winner if it exists. In this paper, the Condorcet correspondence refers to as the following one: if the Condorcet winner exists, it is the unique winner; otherwise, all candidates win.

Our discussion also needs the concept of weak Condorcet winner. Precisely, a candidate in an election is a weak Condorcet winner if and only if it is not beaten by anyone else.

Nearly single-peaked elections. An election  is single-peaked if there is a linear order  of , called an axis, such that for every vote and every three candidates with or , it holds that implies . An election is -axes single-peaked if there are  axes such that every is single-peaked with respect to at least one of . In addition, is a -CP election if there is a -partition of  such that for all , is single-peaked.

Problem formulation. For a voting correspondence , we study the following two problems.

Input:

An election , a distinguished candidate , a multiset  of votes, and a positive integer .

Question:

Is there such that and  uniquely wins with respect to ?

In the above definition, votes in  and  are referred to as registered votes and unregistered votes, respectively. For an instance of CCAV, a subset is called a feasible solution of the instance if  uniquely wins .

Constructive Control by Deleting Votes (CCDV)

Input:

An election , a distinguished candidate , and a positive integer .

Question:

Is there such that and  uniquely wins the election with respect to ?

For an instance of CCAV, a subset is called a feasible solution of the instance if  uniquely wins . An optimal solution of a Yes-instance of CCAV/CCDV refers to as a feasible solution consisting of the minimum votes.

In this paper, we study CCAV and CCDV in -CP (-axes) elections. For CCAV, we mean that is a -CP (-axes) election.

For NP-hardness results, we are only interested in the minimum values of  for which CCAV and CCDV in -CP (-axes) elections are NP-hard. In fact, one can easily show that, for all voting correspondences considered in this paper, if CCAV and CCDV in -CP (resp. -axes) elections are NP-hard, so are they in -CP (resp. -axes) elections. Our NP-hardness results are based on reductions from the following problem.

Restricted Exact Cover by -Sets (RX3C)

Input:

A universe and a collection of -subsets of  such that each  occurs in exactly three elements of .

Question:

Is there an such that and each appears in exactly one element of ?

The RX3C problem is NP-hard [Gonzalez (1985)].

Parameterized complexity. A parameterized problem instance is a tuple , where  denotes the main part and  is a parameter which is often an integer. A parameterized problem is FPT if any of its instance can be determined in time, where  is a computable function and  is the size of the main part. A parameterized problem is para-NP-hard if there is a constant  such that the problem is NP-hard for any parameter greater than . For more detailed introduction to parameterized complexity, we refer to [Downey and Fellows (1992a), Downey and Fellows (1992b)].

## 3 r-Approval

In general elections, CCAV and CCDV for -approval are NP-hard even when  is a constant ( for CCAV and for CCDV) [Lin (2012)]. However, when restricted to single-peaked elections, both problems become polynomial-time solvable (even when  is not a constant) [Faliszewski et al. (2011b)]. We complement these results by first showing that CCAV for -approval in -axes elections is FPT with respect to . Our FPT-algorithm is based on the following two observations

If a vote is single-peaked with respect to an axis , then all approved candidates in the vote lie consecutively in .

For every Yes-instance of the CCAV problem, any optimal solution consists of only unregistered votes approving the distinguished candidate.

The above observations suggest that to solve an instance, we need only to focus on a limited number of candidates—the candidates at most “ far away” from the distinguished candidate  in the -axes of the given instance.

CCAV for -approval in -axes elections is FPT with respect to the combined parameter .

Let  be the components of the input of a CCAV instance as in the definition, where is a -axes election. For each , let  be the score of  with respect to , i.e.,  is the number of votes in  approving . Let  be the multiset of all votes such that . For each vote , let  be the set of candidates ranked in the top- positions, i.e., . Moreover, let . Due to Observation 3, any optimal solution consists of only votes from . Moreover, adding a vote in  never prevents  from winning. Hence, if the given instance is a Yes-instance, there must be a feasible solution consisting of exactly votes. We reset , and seek a feasible solution with  votes in . Obviously, the final score of  is . If there is a candidate such that , the given instance must be a NO-instance. Assume that this is not the case. The question is then whether there are  votes in  such that for every at most of the votes approve . This can be solved in FPT time with respect to

. To this end, we give an integer linear programming (ILP) formulation with the number of variables being bounded by a function of

. We call a vote a -vote if . First, we create for each subset an integer variable  which indicates the number of -votes that are included in the solution. The restrictions are as follows. Let  be the number of -votes in in . First, for each variable , we require that . Second, the sum of all variables should be , i.e., . Third, for each , it must be that . By the result of Lenstra [Lenstra (1983)], this ILP can be solved in FPT time with respect to . Due to Observation 3 contains at most candidates. The theorem follows.

Note that the FPT-algorithm in the proof of Theorem 3 does not need any -axes of the given election. What important is that when the given election is a -axes single-peaked, the cardinality of the set  is bounded from above by . The framework in the proof does not apply to CCDV for -approval in -axes elections. The reason is that any optimal solution of CCDV consists of only votes disapproving the distinguished candidate . Hence, we cannot only confine ourselves to a limited number of candidates.

Now we consider -CP elections. Yang and Guo [Yang and Guo (2017)] developed a polynomial-time algorithm for CCAV for -approval in -peaked elections. As -CP elections are a special case of -peaked elections, their polynomial-time algorithm directly applies to CCAV for -approval in -CP elections. However, if  increases just by one, we show that the problem becomes NP-hard even for . Yang and Guo [Yang and Guo (2017)] also proved that CCAV for -approval in -peaked elections is NP-hard for every . Their proof is via a reduction from the Independent Set on Graphs of Maximum Degree problem and, more importantly, the election constructed in their proof is not a -CP election. We use a completely different reduction to show our result. Particularly, our reduction is from the RX3C problem. The following lemma is easy to see.

Let  be a linear order over  and let be a subset of candidates that are consecutive in . Then we can construct a linear order  over  such that all candidates in  are ranked above all candidates not in , and  is single-peaked with respect to .

CCAV for -approval in -CP elections is NP-hard for every .

Let be an instance of RX3C. We create a CCAV instance with the following components. Consider first .

Candidates . We create in total candidates. In particular, for each , we create a set of four candidates. Let

 C1=⋃cx∈UC(cx)

be the set of all these candidates. Hence, . In addition, for each , we first create three candidates , and  corresponding to , and , respectively, then we create one candidate . Let  be the set of all these candidates corresponding to all . Hence, . Finally, we create a set  of five candidates denoted by , and , respectively. The distinguished candidate is .

We now construct the registered and unregistered votes. For each vote to be created below, we only first specify the approved candidates in the vote, and then we discuss the -axes and use Lemma 3 to specify the linear preference of the vote.

Registered Votes . First, we create  votes approving , and . Then, for each , we create votes approving , and . Finally, for each , there are votes approving , and .

Unregistered Votes . For each , we create four votes as follows:

• approving ;

• approving ;

• approving ; and

• approving .

The above construction clearly takes polynomial time. Now we discuss the preferences of the above votes. Let

 ⊲1=(c11,c21,c31,c41,…,c13κ,c23κ,c33κ,c43κ).

Let  be an order of  such that for every , , all candidates created for  are ordered before all candidates created for . The relative order over the candidates created for each can be any liner order such that the candidates , and  are ranked together. Finally, let . Clearly, for each , the approved candidates restricted to  in each vote lie consecutively on . Then, due to Lemma 3 we can specify the preferences of the votes in a way so that is single-peaked for each .

It remains to prove the correctness.

Assume that is an exact -set cover of . Consider the election after adding the following  votes:

• All  votes  such that ;

• For each , all three votes , and .

As  is an exact -set cover, for each at most one of the above added votes approves . As a result, each candidate has final score at most . Moreover, for each , at most one of the above added votes, corresponding to such that , approves . As a result, each candidate in  has final score at most too. As all unregistered votes approve  but none of , the final score of  is  and the final score of each  is . In summary,  becomes the unique winner in the final election.

The NP-hardness of CCAV for -approval for every can be obtained from the above reduction by adding some dummy candidates. Precisely, for each , we make copies of so that they consecutively lie between and in . For each , we make copies of  and let them lie consecutively with  in . In addition, we create a set  of candidates which consecutively lie on the left side of  in . Finally, we create a set  of  candidates which consecutively lie on the right side of in . Each vote approves the same candidates as defined above together with certain dummy candidates, who do not have any chance to become a winner by adding at most  votes. Precisely, we have the following registered votes.

• For each , votes approving , , , , and the copies of .

• For each , votes approving , , , , and the copies of .

Regarding the unregistered votes, for each , we create four votes respectively approving

• , , , , and all candidates in .

• , , , and all the copies of .

• , , , and all the copies of .

• , , , and all the copies of .

The correctness proof is similar.

Now we turn our attention to CCDV. Yang and Guo [Yang and Guo (2014a)] proved that CCDV for -approval in -peaked elections is NP-hard even for . We strengthen their result by showing that the problem remains NP-hard even when restricted to elections that are both -axes single-peaked and -CP single-peaked. Our reduction is completely different from theirs. In fact, to establish our result, we resort to a property of -regular bipartite graphs which has not been used in the proof of Yang and Guo [Yang and Guo (2014a)]. The -regular bipartite graph in our reduction comes from the graph-representation of the RX3C problem. In general, this property says that for every -regular bipartite graph there are two linear orders over the vertices so that every edge of the graph is between two consecutive vertices in at least one of the two orders. We believe that this property is of independent interest. Recall that -regular graphs are those whose vertices are all of degree .

Let  be a -regular bipartite graph with vertex set  and edge set . Then, there are two linear orders  and  over  and a partition of such that for every and for every edge , it holds that  and  are consecutive in .

We defer the proof of this lemma to Appendix. Figure LABEL:fig-property-regular-graph is an illustrating example.

Now we are ready to unfold the NP-hardness of CCDV for -approval in -axes and -CP single-peaked elections.

For every , CCDV for -approval restricted to elections that are both -axes single-peaked and -CP single-peaked is NP-hard.

Let be an instance of RX3C. We create a CCDV instance with the following components as follows. We first consider and then we discuss how to extend the reduction for any .

Candidates . We create in total candidates. In particular, for each , we create two candidates  and . In addition, we create five candidates denoted by , and , where  is the distinguished candidate. Let

 C1={cix∣i∈{1,2},cx∈U}∪{p,q1,q2,q3,q4}.

Finally, for each , we create three candidates , and  corresponding to , and , respectively. Let  be the set of all candidates corresponding to elements in . Let .

Votes . We only specify here the approved candidates in each created vote, then after the correctness proof we utilize Lemma 3 to specify the linear preferences of all votes so that they are both -axes single-peaked and -CP single-peaked. First, we create one vote approving  and one vote approving . In addition, for each , we create four votes as follows:

• approving ;

• approving ;

• approving ; and

• approving .

It is easy to verify that the winning set is . Precisely, every winning candidate has score  has score , every  where has score , and every candidate in  has score . Finally, we set , i.e., we delete at most  votes.

The above construction clearly takes polynomial time. In the following, we prove the correctness of the reduction.

Assume that is an exact -set cover of . Consider the election after deleting the following votes:

• All  votes  such that ;

• For each , all three votes , and .

Due to the construction and the fact that  is an exact 3-set cover,  has score  and every other candidate has score  after deleting these votes, implying that  becomes the unique winner.

Assume that  is a subset of  with minimal cardinality such that and  becomes the unique winner after deleting all votes in . Due to the minimality of , no vote in  approves . Hence,  has score  after deleting all votes in . Let and . For every , all three votes , where , must be included in , since otherwise one of , and  would have score  after deleting all votes in . Let . It follows from the above analysis that , implying that . On the other hand, as there are candidates in and every vote in  approves two of these candidates, to decrease their scores to at most , we need to delete at least votes, i.e., . This directly implies that and . Let . Clearly, . Due to the above analysis, for every , none of , and  is in , since otherwise there would be more than  votes in . As a result, if there are two  which contain a common element , then  (and ) would have score at least  after the deletion of all votes in , contradicting that  is the unique winner. So, the -subsets in  must be pairwise disjoint, implying that  is an exact -set cover.

Finally, we show that the election constructed above is both a -axes election and a -CP election. We first how that it is -axes single-peaked. To this end, we show that there exist two axes  and  over  such that for every vote constructed above, the approved candidates in the vote are consecutive in at least one of  and . To this end, we need an auxiliary graph. Note that the RX3C instance can be represented by a -regular bipartite graph with vertex-partition . In addition, there is an edge between some and if and only if . Due to Lemma 3, there are two linear orders  and  over  such that for every edge  in the graph where the two vertices  and  are consecutive in one of these two orders. We first construct a linear order  (resp. ) over based on  (resp. ). First, we let  (resp. ) be a copy of  (resp. ) and then we do the following modification.

• For each where , we replace  with the two candidates  and  corresponding to  in  (resp. ). The relative order between  and  in  (resp. ) does not matter.

• For each where , we replace  in  (resp. ) with the three candidates , and  created for the element . The relative order among these three candidates are determined as follows. If  is not the first element in  (resp. ), let , , be the element ordered immediately before  in  (resp. ), i.e.,  and  are consecutive in  (resp. ) and (resp. ). If , we require that  is ordered before everyone in so that the three candidates , , and are consecutive. Symmetrically, if  is not the last element in  (resp. ), and  denotes the element ordered immediately after  in  (resp. ) we have the following requirement: if , we require that  is the last one among , and , so that the three candidates , , and are consecutive. See Figure 2 for an illustration. We order , and  so that the above requirements are fulfilled.

Given the final  and , let

 ⊲1=(q1,q2,p,q3,q4,⊲∗1) and
 ⊲2=(q1,q2,p,q3,q4,⊲∗2).

Clearly, the three candidates , and  are consecutive in both  and , and the three candidates , and  are consecutive in both  and  too. Due to Lemma 3, we can complete the linear order of the vote approving exactly , and  (resp. , and ) so that it is single-peaked with respect to  and, moreover, , and  (resp. , and ) are the top- candidates. Let  be a -subset in . In  and , all the three candidates created for  are consecutive. Let be an element in  and let us consider the vote  whose top- candidates are , and . Clearly,  is an edge in the above mentioned -regular graph. Then, due to Lemma 3 and  are consecutive in at least one of the original orders  and , say, without loss of generality, . Then due to the definition of , the three candidates , and  are consecutive. Therefore, all the three votes created for  can be completed into linear-order votes which are single-peaked with respect at least one of  and  too, and whose top- candidates are exactly those that are approved in these votes. This completes the proof that the constructed election is a -axes election with  and  being the witness.

Now, we show that the above election is also a -CP election. To this end, it suffices to show that is single-peaked for each . Let  be an order of  such that for every , , the two candidates corresponding to