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 NPhard [Bartholdi III et al. (1992), Faliszewski et al. (2011a), Faliszewski et al. (2009), Lin (2012)]. However, when restricted to singlepeaked elections, many of them become polynomialtime solvable [Brandt et al. (2015), Faliszewski et al. (2011b)]. Recall that an election is singlepeaked if there is an order of the candidates, the socalled 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 singlepeaked domain to the general domain with respect to a certain concept of nearly singlepeakedness, where does the complexity of these problems change? A large body of results have been reported with respect to some nearly singlepeaked 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 singlepeaked 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 singlepeaked 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 singlepeaked election. An election is a CP election if there is a partition of the candidates such that the subelection restricted to each is singlepeaked. Clearly, axis elections and CP elections are exactly singlepeaked 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 polynomialtime 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 polynomialtime solvable when restricted to singlepeaked elections. Resolving an open question, Yang [Yang (2017b)] recently proved that CCAV and CCDV for Borda are NPhard even when restricted to singlepeaked 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 singlepeaked width at most and peaked elections were studied. Generally, an election has singlepeaked 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 singlepeaked 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 singlepeaked. Obviously, CP elections are a subclass of peaked elections. In addition, it is known that any election of singlepeaked 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 singlepeaked width [Erdélyi et al. (2017)].
In addition to CCAV and CCDV, many other problems restricted to singlepeaked or nearly singlepeaked 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). ApprovalBased multiwinner voting problems restricted to analogs of singlepeaked 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 singlecrossing and nearly singlecrossing 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 NPhard 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 fixedparameter tractable (FPT) with respect to , but CCDV is already NPhard even for , meaning that CCDV restricted to axes elections is even paraNPhard 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 paraNPhard 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 polynomialtime 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 NPhard for every . We complement these results by showing that determining whether an election is a axes election is polynomialtime solvable, filling the last complexity gap of the problem with respect to .
Constructive Control by Adding Votes (CCAV)  

SP  Axes  CP  CP  general  
approval 
P  FPT (Theorem 3)  P  : P  : P 

: NPhard (Theorem 3)  : NPhard  
Borda 
NPhard  NPhard  
Condorcet 
P  FPT (Theorem 4)  open  NPhard (Theorem 4)  NPhard 
Copeland 
open  NPhard (Theorem 4)  NPhard  NPhard  
Copeland 
P  NPhard (Theorem 4)  NPhard (Theorem 4)  NPhard  
Maximin 
P  NPhard (Theorem 4)  NPhard (Theorem 4)  NPhard 
Constructive Control by Deleting Votes (CCDV)  


SP  Axes  CP  CP  general 
approval 
P  : P  : P  

: NPhard (Theorem 3)  : NPhard  
Borda 
NPhard  NPhard  
Condorcet 
P  FPT (Theorem 4)  open  NPhard (Theorem 4)  NPhard 
Copeland 
open  NPhard (Theorem 4)  NPhard  NPhard  
Copeland 
P  NPhard (Theorem 4)  NPhard (Theorem 4)  NPhard  
Maximin 
P  NPhard (Theorem 4)  NPhard (Theorem 4)  NPhard 
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 nonempty 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 wellknown 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 singlepeaked elections. An election is singlepeaked 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 singlepeaked if there are axes such that every is singlepeaked with respect to at least one of . In addition, is a CP election if there is a partition of such that for all , is singlepeaked.
Problem formulation. For a voting correspondence , we study the following two problems.
Constructive Control by Adding Votes (CCAV) 


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 Yesinstance 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 NPhardness results, we are only interested in the minimum values of for which CCAV and CCDV in CP (axes) elections are NPhard. 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 NPhard, so are they in CP (resp. axes) elections. Our NPhardness 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 NPhard [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 paraNPhard if there is a constant such that the problem is NPhard for any parameter greater than . For more detailed introduction to parameterized complexity, we refer to [Downey and Fellows (1992a), Downey and Fellows (1992b)].
3 Approval
In general elections, CCAV and CCDV for approval are NPhard even when is a constant ( for CCAV and for CCDV) [Lin (2012)]. However, when restricted to singlepeaked elections, both problems become polynomialtime 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 FPTalgorithm is based on the following two observations
If a vote is singlepeaked with respect to an axis , then all approved candidates in the vote lie consecutively in .
For every Yesinstance 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 Yesinstance, 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 NOinstance. 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 FPTalgorithm 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 singlepeaked, 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 polynomialtime algorithm for CCAV for approval in peaked elections. As CP elections are a special case of peaked elections, their polynomialtime algorithm directly applies to CCAV for approval in CP elections. However, if increases just by one, we show that the problem becomes NPhard even for . Yang and Guo [Yang and Guo (2017)] also proved that CCAV for approval in peaked elections is NPhard 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 singlepeaked with respect to .
CCAV for approval in CP elections is NPhard 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
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 , , , .
Finally, we set , i.e., we add at most votes.
The above construction clearly takes polynomial time. Now we discuss the preferences of the above votes. Let
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 singlepeaked 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.
Assume that such that and becomes the unique winner after adding all votes in . As all registered votes disapprove and there are registered votes approving , it must be that . As each unregistered vote approves , the final score of is . Let and be the multiset of the remaining unregistered votes. Then, contains exactly votes in . The reason is as follows. If contains less than votes in , then must contain at least votes in . This implies that there are two votes in which approve a common candidate for some , leading to have a final score at least . This contradicts that is the unique winner. Moreover, if contains some vote where , then none of , , can be included in , since otherwise due to the construction of the votes, one of , , and would have a final score at least , contradicting that is the unique winner. Hence, if contains votes in , then . Hence, implies , a contradiction. Therefore, contains exactly votes in . Let . Due to the above analysis, it holds that . Moreover, for each where , all three votes , , are in (otherwise contains less than votes). From the fact that each candidate in can be approved by at most one vote in , it follows that is an exact set cover.
The NPhardness 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.

votes approving .

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 NPhard even for . We strengthen their result by showing that the problem remains NPhard even when restricted to elections that are both axes singlepeaked and CP singlepeaked. 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 graphrepresentation 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:figpropertyregulargraph is an illustrating example.
Now we are ready to unfold the NPhardness of CCDV for approval in axes and CP singlepeaked elections.
For every , CCDV for approval restricted to elections that are both axes singlepeaked and CP singlepeaked is NPhard.
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
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 singlepeaked and CP singlepeaked. 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 3set 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 singlepeaked. 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 vertexpartition . 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
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 singlepeaked 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 linearorder votes which are singlepeaked 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 singlepeaked for each . Let be an order of such that for every , , the two candidates corresponding to
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