Manipulating Elections by Selecting Issues

11/28/2019 ∙ by Jasper Lu, et al. ∙ Vanderbilt University Nanyang Technological University 0

Constructive election control considers the problem of an adversary who seeks to sway the outcome of an electoral process in order to ensure that their favored candidate wins. We consider the computational problem of constructive election control via issue selection. In this problem, a party decides which political issues to focus on to ensure victory for the favored candidate. We also consider a variation in which the goal is to maximize the number of voters supporting the favored candidate. We present strong negative results, showing, for example, that the latter problem is inapproximable for any constant factor. On the positive side, we show that when issues are binary, the problem becomes tractable in several cases, and admits a 2-approximation in the two-candidate case. Finally, we develop integer programming and heuristic methods for these problems.



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1. Introduction

The study of the extent to which elections are susceptible to subversion by malicious parties has received considerable attention under the general framework of election control. The computational complexity of this problem has been formally studied from a number of perspectives, such as control by adding and deleting candidates and voters Bartholdi et al. (1992); Hemaspaandra et al. (2007), and in the context of different voting systems Menton (2013); Faliszewski et al. (2009b); Erdélyi and Rothe (2010); Hemaspaandra et al. (2009); Erdélyi and Fellows (2010). However, there is an important means of manipulating election outcomes that has been largely ignored: that of determining which issues are discussed and, consequently, which are most salient for voters when they come to the polls.

To illustrate, take three issues, healthcare, environmental regulation, and immigration, and suppose that all voters want universal health coverage and environmental regulation, and a slight majority wish to restrict immigration. Suppose that positions are binary (support or oppose). Now, consider two candidates, one who supports immigration, environmental regulation, universal healthcare, and the second who is opposed to all three. Clearly, if all issues are considered, the former candidate wins in a landslide. However, if one party is able to skew discourse

entirely towards immigration, the second candidate may narrowly win.

We investigate the problem of election control through manipulating issues (which can also be viewed as a novel variant of the bribery problem Faliszewski and Rothe (2016); Put and Faliszewski (2016); Yang et al. (2016)

). In this problem, we assume that voters and candidates can be represented as points in a vector space over issues, where each vector represents one’s (voter’s or candidate’s) position on all issues, and the preference ranking of candidates by a voter is induced by the norm distance between their respective position on issues in the natural way. We then investigate the election control problem in the context of a choice of a subset of issues, whereby the distance between a voter and a candidate in the resulting restricted issue space determines the relative standing of this candidate to others. Our study considers several related variations of this general framework: the decision problem in which the interested party either aspires to have a candidate of their choice win, and the optimization problem of maximizing the support (total number of votes) for a target candidate, all in the context of plurality elections.

We obtain a series of strong negative results. First, we show that not only is the general problem of controlling elections through manipulating issues NP-Hard for both the decision problem and the variant aiming to maximize support, it is actually inapproximable for any constant factor for the latter variant. Moreover, the problem remains hard whether one breaks ties in favor of the target candidate, or not, and even when there is either a single voter, or two candidates. Second, we show that the problem remains hard if we restrict issues to be binary. On the other hand, we observe that under certain restrictions we can obtain positive results. For example, the problem is tractable if there is only a single voter (unlike in the general case), and maximizing support is 2-approximable when there are two candidates. Finally, we provide solution approaches for these problems based on integer linear programming, as well as a greedy heuristic.

1.1. Related Work

Our work is related to two areas of research on social choice: the spatial theory of elections (including lobbying) and election control.

Spatial Theory of Elections and Lobbying Models

Spatial models of elections were first introduced by Hotelling Hotelling (1929), with extensive research following in the decades since (Davis and Hinich, 1966; Enelow and Hinich, 1990; McKelvey and Ordeshook, 1990; Merrill and Groffman, 1999; Anshelevich et al., 2015; Anshelevich and Postl, 2016; Skowron and Elkind, 2017). A major focus areas of research in the spatial model of elections is that of a candidate choosing where to locate in a policy space (Smithies, 1941; Black, 1948). A key development in this field is the Median Voter Theorem (MVT) (Black, 1948), which characterizes the special case of our model with two candidates and one issue. In this case, the winning candidate is the one preferred by the median voter. However, MVT’s assumptions of absolute candidate mobility and global attraction are unrealistic, which continues to stimulate research on this model (Sabato et al., 2017; Shen and Wang, 2017). Algorithmic work in the spatial model has been somewhat more sparse, although with several recent studies focusing largely on social choice functions and distortion relative to a natural social choice function caused by common voting rules, such as plurality (Anshelevich et al., 2015; Anshelevich and Postl, 2016; Skowron and Elkind, 2017).

An important research area within the spatial model is lobbying, whereby an actor wishes to change decisions by voters on issues so that majority vote on each issue corresponds to this actor’s preference Christian et al. (2007); Binkele-Raible et al. (2014). The two clear difference from our proposed research is that in our case, issue preferences determine which candidate wins, rather than votes on each issue separately, and that in our case manipulation targets groups of voters, whereas lobbying research is typically focused on changing votes for a subset of voters. Somewhat related research assumes voters and candidates are fixed actors in a policy space, and considers the game of convincing voters of a candidate’s truthfulness (Austen-Smith, 1990; Lott and Reed, 1989).

Election Control

Election control research focuses on the problem of tampering with an election to either ensure that a candidate wins or loses an election. The spatial theory of elections aims to explain why voters vote the way they do by modeling an election system as sets of voters and candidates as positions in an -dimensional policy space, in which voters vote for those candidates closest to them in Euclidean distance.

The computational problem of constructive election control, in which an adversary manipulates an election to ensure that a candidate wins was first studied by Bartholdi et al. Bartholdi et al. (1992), while Hemaspaandra et al. Hemaspaandra et al. (2007) initiated the study of destructive control. Much work since then has been done in election control under different voting systems, such as range voting (Menton, 2013), approval voting (Erdélyi et al., 2009), and others (Faliszewski et al., 2009b; Erdélyi and Rothe, 2010; Hemaspaandra et al., 2009; Erdélyi and Fellows, 2010), as well as in bribery (Faliszewski et al., 2009a; Erdélyi and Rothe, 2010; Yang et al., 2016; Put and Faliszewski, 2016).

2. Control through Issue Selection

We study the problem of election control through issue selection. To do so, we impose structure on a voting problem by assuming that voter preferences over candidates are solely based on their relative stance on the issues. To be more precise, consider a collection of issues, and a space of possible positions on the issues. Thus, represents a vector of positions on all issues, with the position on (opinion about) issue . In our setting, we have a collection of candidates, , and voters, , where each candidate and voter is characterized by a position vector (representing their respective positions on all issues), which we denote by and , respectively, with . We use () to denote the position of candidate (voter ) on issue , and we refer to the vector of a candidate’s or voter’s beliefs as their belief vector. Denote by the interval of all natural numbers from to , and suppose that voters consider a nonempty subset of issues, , in deciding which candidate to vote for. This set captures those issues which are salient to voters, for example, due to a focus on these during campaigning. We assume that a voter will rank candidates in order of their relative agreement on issues, as captured by an norm for integral with respect to the set of issues . Henceforth, we focus on plurality elections, so that a voter would vote for a candidate which minimizes , where denotes a restriction of to issues in .

We consider two constructive control problems within this framework: control through issue selection (Issue Selection Control (ISC)), and maximizing support (Max Support), which we now define formally.

Definition (Issue Selection Control (ISC)).

Given a set of candidates , voters , and issues, is there a nonempty subset of issues such that a target candidate wins the plurality election?

Definition (Max Support).

Given a set of candidates , voters , and issues, find a nonempty subset of issues which maximizes the number of voters who vote for a target candidate .

For both problems, we must define a rule by which to break ties. We consider both the best-case of undecided voters choosing the target candidate , and the worst-case of undecided voters choosing another candidate. We use the same tie-breaking rule when several candidates are tied.

3. Real-Valued Issues

We begin our study of election control by analyzing its algorithmic hardness when issue positions are unrestricted, i.e., . We show that the problem is computationally intractable, even for a single voter or with only two candidates. However, the problem is tractable when the number of issues is bounded by a constant.

3.1. Issue Selection with a Single Voter

Consider election control through issue selection with only a single voter, , which we term Single-Voter Issue Selection (SVIS). We start by assuming that ties are broken in candidate ’s favor (best-case tie breaking). Note that in this setting, Issue Selection and Max Support are essentially equivalent: in either case, we ask whether there exists a nonempty subset of issues such that when restricted to these issues, candidate wins the voter (with a maximum support of 1 if wins, and 0 if it loses). Equivalently, we ask if there exists a nonempty subset such that


where is the sole voter’s position on issue . Observe that condition (1) holds if and only if

Thus, setting the entries of an auxiliary matrix


we can equivalently ask whether there exists a nonempty subset of the columns of such that the restriction of to these has nonnegative row sums. We will refer to such a restriction of an election as "highlighting" a set of issues.

SVIS with best-case tie breaking is NP-complete for any norm.


First observe that SVIS is in NP. Indeed, given an instance of SVIS and a proposed subset , it is trivial to verify whether satisfies condition (1) in polynomial time.

We now show that SVIS is NP-hard via reduction from 0-1 Integer Linear Programming, which is well-known to be NP-complete. In this problem, we are given a matrix and a vector , and we ask if there exists a vector such that componentwise.

Given an arbitrary instance of 0-1 Integer Linear Programming (ILP), we construct an matrix as follows:

This construction is motivated by the observation that choosing a subset of columns of so that wins the election is analogous to choosing the positions of ones in a vector that satisfies . Each row of corresponds to a candidate belief vector with the constraint vector included as an added issue. We force this issue to be considered by creating a dummy candidate whose beliefs coincide with on all but that issue.

We now construct an instance of SVIS by setting the voter belief vector to be the zero vector and constructing a sequence of candidate belief vectors from .

We do this because we want to arrange that , using positive values of for simplicity. It is then straightforward to see that the original instance of 0-1 Integer Linear Programming is satisfiable if and only if our constructed instance of SVIS is satisfiable, by constructing a 0-1 vector with ones at precisely the indices in , or vice versa. ∎

The worst-case version of SVIS is at least as hard as the best-case version of SVIS.

Proof Sketch.

Consider an matrix representing an arbitrary instance of the best-case version of SVIS and define

where the set . We can create a new matrix as follows:

Recall that in the worst-case version of SVIS, a voter will default to other candidates in cases of a tie. So, we are forced to include issue in in order to win against candidate . Once we include issue , we bias the voter towards the target candidate and against each candidate by a small amount. Because of our choice of , this bias will only affect the election in instances where the candidates are tied. However, we still have to include at least one other issue from to win against candidate .

This construction then turns into the best-case version of SVIS once we begin to consider combinations of issues from with issue . ∎

3.2. Issue Selection with Two Candidates

While issue selection is hard even with a single voter, we now ask whether it remains hard if we have only two candidates. We term the resulting restricted problem Two-Candidate Issue Selection (TCIS). We show that both of the considered problem variants remain NP-hard. Furthermore, Max Support is actually inapproximable to any constant factor even in this restricted setting.

TCIS with best-case tie breaking is NP-complete.


First, observe that TCIS is in NP because, given a set of issues to highlight, we can easily check if wins the election in polynomial time. We use a reduction from 0-1 Integer Linear Programming to prove it’s NP-Hard.

Next, consider the issue selection problem with two candidates and a set of voters . Note that we successfully control the election iff the following condition holds for at least half of the voters (remember that ties are broken in ’s favor):


We now construct a matrix with entries


We can equivalently ask for a nonempty subset of columns of such that the restriction of to those columns maximizes the number of indices s.t. .

Let be our ILP matrix, and - the ILP constraints. Then, we can reduce ILP to TCIS by creating the following matrix :

As in our reduction of SVIS, we represent the constraint vector as an issue that must be put in in order for to win. We also create dummy voters with all negative entries. This will force us to look for assignments of that satisfy all rows that correspond to constraints of ILP. If can win the given election, we return yes for ILP, and no if cannot win.

Finally, we show that for any we can derive voter preferences consistent with it. Since definition of is independent for different issues , it will suffice to do this for an arbitrary issue (th column of , which we denote by ). Consequently, consider a column , and define (the value of with the largest magnitude). Define and . Additionally, define a function for . Clearly, this function is continuous, and while . Then by the intermediate value theorem, for any , we can find a such that . Repeating the process for each issue gives us the construction. ∎

Next, we turn to the Max Support version of the issue selection problem with two candidates; we term this Two-Candidate Max Support (TCMS). We show that not only is it NP-hard, it is inapproximable. TCMS with best-case tie breaking is NP-hard for any norm. Moreover, it cannot be approximated to any constant factor unless .


We can now show that TCMS is NP-hard by restricting to and reducing from Maximum Independent Set (MIS). Given an undirected graph on vertices, MIS asks to select a maximal subset of vertices so that is an independent set (i.e., no pair of vertices in is connected by an edge).

Given any instance of MIS, we can represent that instance as an instance of TCMS by first creating a matrix with every value along the diagonal equal to . For every pair of vertices , set if and are connected in , and otherwise. Now, if we were to select an issue corresponding to vertex with neighbor , then we cannot hope to select any other subset of issues such that row sums to greater than or equal to 0. Thus, the action of selecting columns of to include in corresponds to selecting vertices of to be in our independent set, and maximizing the number of rows in this manner corresponds to finding a maximum independent set.

To complete the reduction, what remains to prove is that we can derive voter belief and candidate belief vectors for any constructed in this manner. The associated lemma is provided in the supplement.

Inapproximability follows directly from our reduction of MIS to TCMS: we know that MIS is NP-hard to approximate within any constant factor Arora and Barak (2009), and our reduction from MIS is approximation-preserving. ∎

The next results show that the worst-case tie breaking setting is no easier than when ties are broken in ’s favor. The worst-case version of TCIS is at least as hard as the best-case version of TCIS for the two-candidate case.

Proof Sketch.

Given an matrix associated with a two-candidate instance of best-case issue selection, define as in the proof of Theorem 3.1. Further, we let , and create a matrix as follows:


Once again, we choose a value of such that will affect the election only if a voter is undecided. The proper assignment is shown in the supplement.

Recall that in the worst-case version of TCIS, undecided voters (rows of with a net zero value) will default to a candidate other than . With the addition of column , any undecided voters will now be “nudged” in the direction of instead. Also, since the values of column are smaller than the difference of any two values of , the issue affects the election only if a voter is actually undecided. So, issue appropriately mimics the weak inequality used in the best-case version of TCIS, and if a candidate wins an election in the worst-case reduction, they win the election in the best-case version, and vice versa.

Note: we add extra voters to the problem to set things up such that including issue would not be sufficient for winning the election. We also choose voters specifically so that we can be guaranteed to split voters evenly between and with our assignments in Equations 5 and  6. ∎


The worst-case version of TCMS is NP-hard.

4. Binary Issues

We have shown that election control through issue selection is hard in general. However, real world opinions may have a variety of restrictions. For example, legislative issues can be viewed as binary issues, where a voter opinion can take only two values: support or oppose.

Formally, in binary versions of the issue selection problems, . Voters vote for the candidate with whom they agree on most issues. Let Binary Issue Selection Control (BISC) be the variant of Issue Selectionover a binary domain and, similarly, let Binary Max Support (BMS) be the corresponding variant of the Max Support problem.

4.1. Binary Issue Selection with 1, 2 and 3 Voters

We start by considering again the problem of issue selection with a single voter, which we showed to be NP-Hard in the general case of real-valued issues. We show that this problem is now in P.

As before, it suffices to consider solely Single-Voter BISC . We start with the case when ties are broken in ’s favor (best-case tie-breaking). Consider the following Single Issue Win algorithm:

Check if there is an issue such that either (a) agrees with the voter , or (b) no other candidate agrees with . If it exists, return YES. Otherwise, return NO.

The Single Issue Win algorithm solves Single-Voter BISC with best-case tie-breaking.


It suffices to show that whenever Single Issue Win returns NO, cannot win the election. Consider an arbitrary subset of issues . Since the answer is NO, it must be that for each issue , disagrees with on . Consequently, . Choose a which agrees with on some issue . Then , that is, cannot win for issues restricted to . Since is arbitrary, the result follows. ∎

In fact, we can easily generalize the algorithm for a single voter to a setting with two voters by simply applying the algorithm for each voter.


2-Voter BISC problem with best-case tie-breaking is poly-time solvable.

Next, we show that the problem is in P for one and two voters even with worst-case tie-breaking, although the algorithmic approach is quite different. For worst-case tie-breaking, we propose the following Agree On Issues algorithm:

Let be the set of all issues on which agrees with . If wins over each other candidate when issues are restricted to , return YES. Otherwise, return NO.

The Agree On Issues algorithm solves Single-Voter BISC with worst-case tie-breaking.


It suffices to consider the case when we return NO. Suppose there is some that wins when we restrict to . Then it must be that also agrees with on all issues in (and any subset thereof). Consider an arbitrary subset of issues , and let if agrees with on issue . ’s difference from is then . Since the difference between and is , the result follows. ∎

The same approach is also applicable to 2-Voter BIS.


2-Voter BISC with the wost-case tie-breaking is poly-time solvable.


For the candidate to win, both voters must support her. Without loss of generality, we can assume that opinion on all isues is . Let be the set of all issues on which agrees with both voter and . Similarly to Theorem 4.1, if does not win against each other candidate over the set , then no other subset of issues will achieve ’s win. ∎

Remarkably, while BSIC with 1 and 2 voters are efficiently solvable for both best-case and worst-case tie-breaking, with 3 voters we see a qualitative difference in complexity, depending on how ties are broken. First, we observe that the 3-voter case with worst-case tie-breaking is tractable.


3-Voter Binary Issue Selection with the worst-case tie-breaking is poly-time solvable.


By Corollary 4.1 we can test in poly-time whether any given pair of voters can be won over by . Applying this to each of the three possible pairs of voters, we can determine in poly-time whether the support of any two voters can be obtained simultaneously. If so, then can be made to win. Otherwise no subset of issues will make the winner. ∎

Now, we show that the problem becomes hard with best-case tie-breaking even with only 3 voters. 3-Voter Binary Issue Selection with the best-case tie-breaking is NP-hard.


The proof relies on a reduction from the Exact 3-Cover (X3C) problem. An instance of X3C is governed by – number of elements, – the number of sets. In the reduced instance we will denote by the preferred candidate (and assume that his opinion on all issues is ), – the candidate whose opinion on every issue is (zero), – the voter whose opinion on every issue is . This implies that to win the election should gain the support of both voters and . In addition we will denote by the number of issues in the reduced instance, setting it to . Finally, we will set the number of candidates to and name them .

The preferences of and over the issues are as follows:

Preferences of candidates take a more complex form

  • For issues from through . These preferences will encode the X3C instance. In particular, candidates will have opinion on the issue if and only if the set in the X3C instance is . Otherwise the opinion of these three candidates on the issue will be (zero).

  • On issues through all candidates have (zero) opinion.

  • On the issue all candidates have opinion (zero)

  • On issue all candidates have opinion

  • Candidate has opinion on issues and opinion (zero) on the issues and

  • Candidates has opinions in the complete opposion to candidate

Let us now show that if we have a solution to the resulting Issue Selection Control problem, we can recover a solution for the original X3C instance.

Candidate , with all his opinions set to (zero), serves as a kind of reference for voters. Thus, given a selection of issues, the preferred candidate will gain the support of a voter only if they agree on at least as many issues in as they disagree. As a result, Issue Selectionsolution should contain equal number, , of issues from the set and from the set . Consequently, candidate will agree with any voter on exactly issues.

Notice that both the issue and must be selected in a solution to the Issue Selection Control. To see this consider the follwoing two cases

  • Neither , nor are in the solution set, of issues. Still, an equal number of elements (denoted earlier by ) must be selected from the sets of issues and for the solution set . Wlog., issue . Then voter agreed with the candidate on issues ( issues from the set and issue ). As a result, voter would not vote for candidate . Thus , that does not contain neither nor , can not be a valid solution to our BISC instance.

  • Only one among issues and is selected as a part of the solution set of issues . If it is the issue , then voter agreed with the candidate on issues and with candidate on issues only. Thus, would not vote for , and is not a valid soluion. Similarly, if was selected, then candidate will win the support of , once again preventing from winning.

Now, with both issues and chosen, let us show how we can obtain a solution to the original X3C problem from the solution set of issues to the reduced BISC problem. The set of issues makes candidate the winner of the election. Let . We will show that the collection is the solution to the original X3C instance.

  1. If there is an element that belongs to two different sets in the collection , then agrees with on 2 issues from and on issues from . Totalling agreements between and . Which implies that will not vote for , and contradicts being the winner.

  2. If there exists an element that does not belong to any set in the collection , then for all . As a consequence agrees with on issues from the set of issues and on both issues and . This totals agreements between and , entailing that will not vote for , contradicting being the winner.

As a result, the collection constructed from the BISC solution is a proper solution to the original X3C instance, i.e. every element belong to 1 and only 1 set.

Let us now show that a solution to the X3C instance can be translated into a solution to the BISC reduction instance.

Let be a legal solution to the X3C instance. Then set the selection of issues . Notice that is the number of elements in the X3C instance, and therefore and .

By the choice of , it must hold that agrees with every candidate once on issues and times on issues . Overall and agree on issues. Candidate agrees with on issues only, totalling agreements as well. Similarly, candidates and rake in agreements. Thus, by the tie-breaking rule, votes for .

Similarly, is matched with the opinion of over issues from the set and 2 more matches are produced over issues . This totals matches between and . Similarly to , also agrees with and on issues. Again, tie-breaking will decide in favour of . Thus has the support of both and and becomes the winner.

We conclude that the original X3C instance has a solution if and only if the reduction instance of BISC has a solution. ∎

4.2. Binary Issue Selection with Two Candidates

With an arbitrary number of voters and only two candidates, even the BISC problem with best-case tie-breaking is hard. With two candidates, BISC with best-case tie-breaking is NP-complete.


It is evident that BISC problem is in NP, so we only need to show that it is NP-hard. We will do so by a reduction from Hitting Set, where denotes the number of elements, – the number of sets, and — the number of elements which should be chosen as the hitting set. We construct a profile for BISC problem with 2 candidates, issues and voters, where is such that and .

We assume that the preferred candidate is and set his opinion to on all issues. All opinions of his rival, , are set to (zero). We then arrange voters into blocks, as follows:

  • [Block 1.] Two voters. The first one has opinion (zero) for issues from through issue , and opinion for issues from to . The second voter has an opposite opinion wrt all issues.

  • [Block 2.] Second block consists of voters divided into k sub-blocks. For every sub-block, opinions of voters on issues from to encode the hitting set problem instance. That is, voter has opinion 1 on issue if and only if element for all . For issues from to , all voters of the sub-block will have the same (zero) opinion on issue and on all other issues.

  • [Block 3.] This block consists of voters whose opinion on all issues is 0.

Let us now show the correctness of this reduction. Let be a set issues chosen to make the winner. Consider voters who support . Evidently, nobody from Block 3 is among them — no matter which issues were chosen, voters from Block 3 will support . As a result, has at least votes. Hence, all voters from Blocks 1&2 should vote for to make him the winner.

Consider voters in Block 1. They both vote for , therefore, consists of equal number of elements from both issue sets and . Otherwise, there are (w.l.o.g.) more issues from than from . Which implies that the second voter from Block-1 has more negative () opinions than positive (), and he will vote for the candidate . Additionally that means at most issues were picked from both sets. Denote this number by .

W.l.o.g. issue is chosen from the set . Thus, voters from the first sub-block of Block-2 have ’s and one as an opinion on issues in . Therefore, all voters from this sub-block should have at least one positive () opinion on issues chosen from the issues set . That is, these issues represent a hitting set with elements where .

Similarly a solution for the BISC can be constructed from a given Hitting Set solution.

This proof is easy to adapt to worst-case tie-breaking. ∎


The BMS problem is NP-hard.

Although Binary Max Support is NP-hard, we now show that it is easy to achieve a -approximation using the following Best-Single-Issue algorithm: choose one issue that maximizes the net number of voters captures. The Best-Single-Issue algorithm approximates -candidate Binary Max Support to within a factor of , for best-case and worst-case tie-breaking.


Let’s denote number of voters by and the number of issues by . Among two candidates and the promoted one is . Without loss of generality, we can assume that candidate has opinion 1 on every issue. We will provide proof for the case of best-case tie-breaking and will describe changes needed to transform this proof into proof for worst-case tie-breaking.

  • [Case 0.] There is an issue s.t. candidate also has opinion 1 about this issue. Therefore, if we highlight only this issue all voters will vote for because of tie-breaking. Thus, that is an optimal solution (and as such approximation within factor 2 of optimal solution). It is also the issue that captured the greatest number of voters for if highlighted. From now on we can assume that opinion of candidate is 0 for all issues.

  • [Case 1.] There exists an issue s.t. at at least voters have same opinion as . If highlighted such issue will capture for at least voters. That is, for issue that causes to capture the greatest number of voters it is at least voters too. Thus, it is provide -approximation, because optimal solution is at most .

  • [Case 2.] Now we can assume that for all issues less then voters have opinion 1. Denote the largest such number by and show that is -approximation of optimum. Assume the contrary. W.l.o.g. issues maximizes support for candidate . By choice of the number of opinions which equals to 1 over all issues is at most . On the other hand voter supports candidate if and only if he has opinion 1 for at least issues among . By assumption there are strictly more than such issues. That is, on issues opinion 1 shared strictly more than times. Obtained contradiction proves the theorem.

This proof can be easily adopted for the case of worst-case tie breaking. It is easy to see that if candidate has opinion 1 on all issues then for every highlighted set of issues support of candidate will be 0. Thus, any single issue provides -approximation of optimum. Therefore, we may assume that there exist issue on which candidate has opinion 0.

Evidently, if there is optimum such that on some of highlighted issues candidate has opinion 1. W.l.o.g. this issue then is also optimum. Therefore, we may assume that candidates have different opinions on all issues. Thus, it is enough to consider cases 1 and 2. The proof for case 1 remains unchanged. For case 2 we should change the counting of number of points needed to obtain at least votes in favor of candidate . A voter would only vote for if he has opinion 1 for issues . Therefore, the number of opinions 1 is , yielding same contradiction as in best-case tie-breaking. ∎

5. Algorithmic Approaches

We now present several general algorithmic approaches for Max Support: 1) exact approaches based on integer linear programming (ILP), and 2) a heuristic approach which works well in practice.

Integer Linear Programming: Define as follows:


Define . The following ILP computes an optimal solution for (best-case) Max Support: equationparentequation


Constraint (8b), ensures that iff is the most favored by voter . A similar approach can be used to develop a ILP approach for the Issue Selection Control problem.

Greedy Heuristic: Finally, we present a simple greedy algorithm for the Max Support problem, where we iteratively add one issue at a time that maximizes the net gain in voters. We stop by adding any single issue would decrease the number of voters captured.

6. Experiments

We now compare the performance of our exact and heuristic solution algorithms for the binary and continuous versions of the issue selection problem. We consider the greedy heuristics described above, as well as Best-Single-Issue.

We run all of our experiments assuming a worst-case tie-breaking rule and generate random synthetic test cases. For continuous test problems, we sample candidate and voter belief vectors from the multivariate normal distribution with a mean of

and a random covariance matrix. A similar generative model for Boolean issues, tends to produce problem instances in which Best-Single-Issue is nearly always optimal. Consequently, we generate a more specialized distribution of these instances as follows. We first construct a vertex-weighted complete binary tree on vertices. Each vertex is assigned an independent random weight

drawn from the uniform distribution on

. To produce a sample from , we perform a directed random walk from its root to one of its leaves. The sequence ( for left movements, and for right) emitted by this process is then the desired sample from .

We default to candidates, voters, and issues. To generate each plot, we fix of these parameters and vary the rd. We generate instances of Max Support for each set of parameter values, and run the heuristics on the instances. The plotted values are averages of the ratio of the number of voters captured and the optimal solution.

Figure 1. Plots of experimentally observed approximation ratios as functions of the numbers of candidates, voters, and issues in synthetic test cases for binary (left) and continuous (right) versions of Max Support.

We find that for most instances of Max Support with binary issues, our greedy heuristic does not significantly outperform Best-Single-Issue in the two-candidate setting as number of issues and voters increase. This is because the number of instances in which a combination of issues can get us more voters than a single best issue is increasingly unlikely. However, the greedy algorithm outperforms Best-Single-Issue on instances of Binary Max Support with greater than candidates. We can also observe that on the specific distribution of binary issue instances we generate, the quality of heuristic solutions degrades rapidly with the number of candidates.

We find that for Max Support with real-valued issues, the greedy algorithm significantly outperforms Best-Single-Issue. For a small number of candidates (), the greedy algorithm seems to perform within of optimal. Interestingly, as the number of voters increases, the greedy algorithm improves in quality on our randomly generated problem instances. In all cases, we can also observe that the heuristics tend to be close to optimal.

7. Conclusion

When candidates participate in an election, they must choose policies and issues to stress in their campaigns. We introduce and study the problem of election control through issue selection. We find a number of strong negative results for the problem, and show that, even though we canno t provide formal approximation guarantees for a continuous instance of Max Support, a simple greedy heuristic performs well. Moreover, restricting issues to be binary admits further positive results, including a -approximation.

This work was in part supported by NSF (IIS-1526860, IIS-1905558), NTU SUG M4081985 and MOE AcRF-T1-RG23/18 grants.


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