1 Introduction
We study how multiwinner voting rules (i.e., procedures used to select fixedsize committees of candidates) react to (small) changes in the input votes. We are interested both in the complexity of computing the smallest modification of the votes that affects the election outcome, and in the extent of the possible changes. We start by discussing our ideas informally in the following example.
Consider a researchfunding agency that needs to choose which of the submitted project proposals to support. The agency asks a group of experts to evaluate the proposals and to rank them from the best to the worst one. Then, the agency uses some formal process—here modeled as a multiwinner voting rule—to aggregate these rankings and to select projects to be funded. Let us imagine that one of the experts realized that instead of ranking some proposal as better than , he or she should have given the opposite opinion. What are the consequences of such a “mistake” of the expert? It may not affect the results at all, or it may cause only a minor change: Perhaps proposal would be dropped (to the benefit of or some other proposal) or would be selected (at the expense of or some other proposal). We show that while this indeed would be the case under a number of multiwinner voting rules (e.g., under the Borda rule; see Section 2 for definitions), there exist other rules (e.g., STV or the Chamberlin–Courant rule) for which such a single swap could lead to selecting a completely disjoint set of proposals. The agency would prefer to avoid situations where small changes in the experts’ opinions lead to (possibly large) changes in the outcomes; so the agency would want to be able to compute the smallest number of swaps that would change the result. In cases where this number is too small, the agency might invite more experts to gain confidence in the results.
More formally, a multiwinner voting rule is a function that, given a set of rankings of the candidates and an integer , outputs a family of size subsets of the candidates (the winning committees). We consider the following three issues (for simplicity, below we assume to always have a unique winning committee):

We say that a multiwinner rule is robust if (1) swapping two adjacent candidates in a single vote can lead to replacing no more than candidates in the winning committee^{1}^{1}1Indeed, the formal definition is more complex due to taking care of ties., and (2) there are examples where exactly candidates are indeed replaced; we refer to as the robustness level of . Notably, the robustness level is between and , with robust being the strongest form of robustness one could ask for. We ask for the robustness levels of several multiwinner rules.

We say that the robustness radius of an election (for committee size ) under a multiwinner rule is the smallest number of swaps (of adjacent candidates) which are necessary to change the election outcome. We ask for the complexity of computing the robustness radius (referred to as the Robustness Radius problem) under a number of multiwinner rules (this problem is strongly related to the Margin of Victory [19, 6, 25, 4] and Destructive Swap Bribery problems [12, 24]; in particular, it follows up on the study of robustness of singlewinner rules of Shiryaev et al. [24])

We ask how many random swaps of adjacent candidates are necessary, on average, to move from a randomly generated election to one with a different outcome. Doing experiments, we assess the practical robustness of our rules.
There is quite a number of multiwinner rules. We consider only several of them, selected to represent a varied set of ideas from the literature (ranging from variants of scoring rules, through rules inspired by the Condorcet criterion, to the eliminationbased STV rule). We find that all these rules are either robust (so a single swap can replace at most one committee member) or are robust (so a single swap can replace the whole committee of size ).^{2}^{2}2We also construct somewhat artificial rules with robustness levels between and . Somewhat surprisingly, this phenomenon is deeply connected to the complexity of winner determination. Specifically, we show (under mild assumptions) that if a rule has a constant robustness level, then it has a polynomialtime computable refinement (that is, it is possible to compute one of its outcomes in polynomial time). Since for many rules the problem of computing such a refinement is hard, we get a quick way of finding out that such rules have nonconstant robustness levels.
The Robustness Radius problem tends to be hard (sometimes even for a single swap) and, thus, we seek fixedparameter tractability (FPT) results. For example, we find several FPT algorithms parametrized by the number of voters (useful, e.g., for scenarios with few experts, such as our introductory example). See Table 1 for an overview on our theoretical results.
We complement our work with an experimental evaluation of how robust are our rules with respect to random swaps. On the average, to change the outcome of an election, one needs to make the most swaps under the Borda rule.
Voting Rule  Robustness Level  Complexity of Robustness Radius 

SNTV, Bloc, Borda ()  
Copeland ()  hard, FPT()  
NED (hard [1])  hard, FPT()  
STV (hard^{3}^{3}3For STV there is a polynomialtime algorithm for computing a single winning committee, but deciding if a given committee wins is hard. [9])  hard(B), FPT(), FPT()  
(hard [22, 18, 3])  hard(B), FPT(), FPT() 
2 Preliminaries
Elections. An election consists of a set of candidates and of a collection of voters . We consider the ordinal election model, where each voter is associated with a preference order , that is, with a ranking of the candidates from the most to the least desirable one (according to this voter). A multiwinner voting rule is a function that, given an election and a committee size , outputs a set of size subsets of , referred to as the winning committees (each of these committees ties for victory).
(Committee) Scoring Rules. Given a voter and a candidate , by we denote the position of in ’s preference order (the topranked candidate has position and the following candidate has position , and so on). A scoring function for candidates is a function that associates each candidateposition with a score. Examples of scoring functions include (1) the Borda scoring functions, ; and (2) the Approval scoring functions, defined so that if and otherwise ( is typically referred to as the Plurality scoring function). For a scoring function , the score of a candidate in an candidate election is defined as
For a given election and a committee size , the SNTV score of a size committee is defined as the sum of the Plurality scores of its members. The SNTV rule outputs the committee(s) with the highest score (i.e., the rule outputs the committees that consist of candidates with the highest plurality scores; there may be more than one such committee due to ties). Bloc and Borda rules are defined analogously, but using Approval and Borda scoring functions, respectively. The Chamberlin–Courant rule [7] (abbreviated as ) also outputs the committees with the highest score, but computes these scores in a somewhat different way: The score of committee in a vote is the Borda score of the highestranked member of (the score of a committee is the sum of the scores from all voters).
SNTV, Bloc, Borda, and are examples of committee scoring rules [11, 13]. However, while the first three rules are polynomialtime computable, winner determination for is wellknown to be hard [22, 18, 3].
CondorcetInspired Rules. A candidate is a Condorcet winner (resp. a weak Condorcet winner) if for each candidate , more than (at least) half of the voters prefer to . In the multiwinner case, a committee is Gehrlein stronglystable (resp. weaklystable) if every committee member is preferred to every nonmember by more than (at least) half of the voters [14], and a multiwinner rule is Gehrlein stronglystable (resp. weaklystable) if it outputs exactly the Gehrlein stronglystable (weaklystable) committees whenever they exist. For example, let the NED score of a committee be the number of pairs such that (i) is a candidate in , (ii) is a candidate outside of , and (iii) at least half of the voters prefer to . Then, the NED rule [8], defined to output the committees with the highest NED score, is Gehrlein weaklystable. In contrast, the Copeland rule is Gehrlein stronglystable but not weaklystable (the Copeland score of a candidate , where , is the number of candidates such that a majority of the voters prefer to , plus times the number of candidates such that exactly half of the voters prefer to ; winning Copeland committees consist of candidates with the highest scores). Detailed studies of Gehrlein stability mostly focused on the weak variant of the notion [2, 16]. Very recent findings, as well as results from this paper, suggest that the strong variant is more appealing [1, 23]; for example, all Gehrlein weaklystable rules are hard to compute [1], whereas there are stronglystable rules (such as Copeland) that are polynomialtime computable.
STV. For an election with candidates, the STV rule executes up to rounds as follows. In a single round, it checks whether there is a candidate who is ranked first by at least voters and, if so, then it (i) includes into the winning committee, (ii) removes exactly voters that rank first from the election, and (iii) removes from the remaining preference orders. If such a candidate does not exist, then a candidate that is ranked first by the fewest voters is removed. Note that this description does not specify which voters to remove or which candidate to remove if there is more than one that is ranked first by the fewest voters. We adopt the paralleluniverses tiebreaking model and we say that a committee wins under STV if there is any way of breaking such internal ties that leads to him or her being elected [9].
We can compute some STV winning committee by breaking the internal ties in some arbitrary way, but it is hard to decide if a given committee wins [9].
Parametrized Complexity. We assume familiarity with basic notions of parametrized complexity, such as parametrized problems, algorithms, and hardness. For details, we refer to the textbook of Cygan et al. [10].
In a couple of proofs in the paper we use the concept of the dissatisfaction of the voters from a committee according to the rule. The dissatisfaction of a voter from a committee is equal to minus the Borda score of the most preferred member of (according to ). Thus, if contains ’s top preferred candidate, then the dissatisfaction of from is equal to zero. If contains ’s second most preferred candidate, then such dissatisfaction is equal to one, etc. The total dissatisfaction of the voters from is defined as the sum of the dissatisfactions over all individual voters.
3 Robustness Levels of Multiwinner Rules
In this section we identify the robustness levels of our multiwinner rules. We start by defining this notion formally; note that the definition below has to take into account that a voting rule can output several tied committees.
Definition 1.
The robustness level of a multiwinner rule for elections with candidates and committee size is the smallest value such that for each election with , each election obtained from by making a single swap of adjacent candidates in a single vote, and each committee , there exists a committee such that .
All rules that we consider belong to one of two extremes: Either they are robust (i.e., they are very robust) or they are robust (i.e., they are possibly very nonrobust). We start with a large class of rules that are robust.
Proposition 1.
Let be a voting rule that assigns points to candidates and selects those with the highest scores. If a single swap in an election affects the scores of at most two candidates (decreases the score of one and increases the score of the other), then the robustness level of is equal to one.
Proof.
Let be an election, be a committee size, and be committee in . We write to denote the individual score of a candidate in . We rename the candidates so that (i) and (ii) . Now consider an election obtained from with a single swap. This swap can increase the score of at most one candidate, say , while decreasing the score of at most one other candidate, say . There are several cases to consider:

If and , then is still winning in .

If and , then either or is winning in .

If and , then either or is winning in .

If and , then either or or or is winning in .
In each case, there is a committee such that . ∎
Proposition 1 This suffices to deal with four of our rules.
Corollary 1.
SNTV, Bloc, Borda, and Copeland (for each ) are robust.
In contrast, Gehrlein weaklystable rules are robust.
Proposition 2.
The robustness level of each Gehrlein weaklystable rule is .
Proof.
Consider the following election, described through its majority graph (in a majority graph, each candidate is a vertex and there is a directed arc from candidate to candidate if more than half of the voters prefer to ; the classic McGarvey’s theorem says that each majority graph can be implemented with polynomially many votes [21]). We form an election with candidate set , where and , and with the following majority graph: The candidates in form one cycle, the candidates in form another cycle, and there are no other arcs (i.e., for all other pairs of candidates the same number of voters prefers to as the other way round). We further assume that there is a vote, call it , where is ranked directly below (McGarvey’s theorem easily accommodates this need).
In the constructed election, there are two Gehrlein weaklystable committees, and . To see this, note that if a Gehrlein weaklystable contains some then it must also contain all other members of (otherwise there would be a candidate outside of the committee that is preferred by a majority of the voters to a committee member). An analogous argument holds for .
If we push ahead of in vote , then a majority of the voters prefers to . Thus, is no longer Gehrlein weaklystable and becomes the unique winning committee. Since (1) and are disjoint, (2) is among the winning committees prior to the swap, and (3) is the unique winning committee after the swap; we have that every Gehrlein weaklystable rule is robust. ∎
To complete this section we show that the robustness levels of CC and STV are . Such negative results seem unavoidable among rules that—like CC and STV—provide diversity or proportionality (we discuss this further in Section 4).
Proposition 3.
The robustness level of is .
Proof.
We construct the following election with the candidate set , , . We construct one vote as follows:
and for each we construct two votes as follows:
Observe that the only winning committee is . To see this, notice that has dissatisfaction of only (the concept of dissatisfaction is introduced at the beginning of the appendix). Further, every voter has a different favorite candidate, there are voters but the committee size is only . Hence, every at least equally good committee must contain of the “favorite” candidates from and every voter that is not represented by her favorite candidate must be represented by her second choice. Now, without candidate being in the committee, voter cannot be represented by its second choice because . So, is indeed in the committee and so the remaining committee members are from since only the voters , can be represented by their second choices.
If we swap and in the first vote, then, following analogous argumentation, the unique winning committee becomes . Finally, we mention that the construction above works for any committee size. ∎
Proposition 4.
The robustness level of STV is .
Proof.
Let us fix a committee size and consider the set of candidates , where and . We construct the following profile. We start by taking the votes representing each of the possible rankings over , and remove these votes which rank or first. Next, for each we add voters who rank first, second, next all other candidates from ranked in some fixed arbitrary order, followed by the remaining candidates from . Similarly, for each we add voters who rank first, second, next all other candidates from , and finally, the remaining candidates from ; additionally, in one of these votes, call it the pivotal vote, we swap and ( is chosen arbitrarily).
In the profile described above, STV eliminates first. After is removed, each candidate from will be ranked top by (significantly) more voters than each candidate from . As a result, in each of the next steps candidates from will be eliminated. Thus, is the unique winner in this profile. Now, let us analyze what happens when we swap and in the pivotal vote. STV might eliminate first. By the same reasoning as before, we infer that is a possible winner in such adjacent profile. The claim now follows since . ∎
One may wonder whether there exist any voting rules with robustness level between and . Although we could not identify any classical rules with this property, we found natural hybrid multistage rules which satisfy it. For example, the rule which first elects half of the committee as Borda does and then the other half as CC does has robustness level of roughly .
Proposition 5.
For each , there is an robust rule.
Proof.
We consider the following voting rule called BordaCC which first selects committee members exactly as Borda would do, and then selects further candidates that maximize the CC score of the whole committee.
As for the upper bound , observe that we can reuse the arguments from the proof of Proposition 1. To this end, let be an election and let be an election obtained from by a single swap of adjacent candidates. Let be a winning committee in . Following the proof of Proposition 1 there is some being a winning committee in with (even if all additional members selected during the respective CCphases are pairwise disjoint).
For the lower bound, consider the following election which is an extension of the election used in Proposition 3. We construct an election with the candidate set , , , . We construct one vote as follows:
for each we construct two votes as follows:
and for each we construct copies of the following two votes:
Clearly, in the first phase all candidates from will be selected which ensures that all voters are perfectly satisfied. Furthermore, candidate will be selected as candidate with the th highest Borda score. Hence, one can verify that, similarly as normal CC does on the election used in the proof of Proposition 3, our rule selects all candidates from as the remaining committee members. Furthermore, by swapping candidate and candidate in the first vote , candidate will be selected instead of candidate as candidate with the th highest Borda score in the first phase and all candidates from will be selected as the remaining committee members in the second phase. Summarizing, the unique winning committee for this election is . If we, however, swap and in the first vote, then, the unique winning committee becomes .
∎
4 Computing Refinements of Robust Rules
It turns out that the dichotomy between robust and robust rules is strongly connected to the one between polynomialtime computable rules and those that are hard. To make this claim formal, we need the following definition.
Definition 2.
A multiwinner rule is scoringefficient if the following holds:

For each three positive integers , and () there is a polynomialtime computable election with voters and candidates, such that at least one member of can be computed in polynomial time.

There is a polynomialtime computable function that for each election , committee size , and committee , associates score with , so that consists exactly of the committees with the highest score.
The first condition from Definition 2 is satisfied, e.g., by weakly unanimous rules.
Definition 3 (Elkind et al. [11]).
A rule is weakly unanimous if for each election and each committee size , if each voter ranks the same set of candidates on top (possibly in different order), then .
All voting rules which we consider in this paper are weakly unanimous (indeed, voting rules which are not weakly unanimous are somewhat “suspicious”). Further, all our rules, except STV, satisfy the second condition from Definition 2. For example, while winner determination for CC is indeed NPhard, computing the score of a given committee can be done in polynomial time. We are ready to state and prove the main theorem of this section.
Theorem 6.
Let be a robust scoringefficient multiwinner rule. Then there is a polynomialtime computable rule such that for each election and committee size we have .
Proof.
Our proof proceeds by showing a polynomialtime algorithm that, given an election and committee size , finds a single committee such that ; we define to output .
Let be our input election and let be the size of the desired committee. Let be an election with , whose existence is guaranteed by the first condition of Definition 2, and let be a size winning committee for this election, also guaranteed by Definition 2.
Let be a sequence of elections such that , , and for each integer , we obtain from by (i) finding a voter and two candidates and such that in voter ranks right ahead of , but in voter ranks ahead of (although not necessarily right ahead of ), and (ii) swapping and in ’s preference order. We note that at most swaps suffice to transform into (i.e., ).
For each , we find a committee . We start with (which satisfies our condition) and for each , we obtain from as follows: Since is robust, we know that at least one committee from the set is winning in . We try each committee from this set (there are polynomially many such committees) and we compute the score of each of them (recall Condition 2 of Definition 2). The committee with the highest score must be winning in and we set to be this committee (by Definition 2, computing the scores is a polynomialtime task).
Finally, we output . By our arguments, we have that . Clearly, our procedure runs in polynomial time.
∎
Theorem 6 generalizes to the case of robust rules for constant : Our algorithm simply has to try more (but still polynomially many) committees .
Let us note how Theorem 6 relates to singlewinner rules (that can be seen as multiwinner rules for ). All such rules are robust, but for those with hard winner determination, even computing the candidates’ scores is hard (see, e.g., the survey of Caragiannis et al. [5]), so Theorem 6 does not apply.
5 Complexity of Computing the Robustness Radius
In the Robustness Radius problem we ask if it is possible to change the election result by performing a given number of swaps of adjacent candidates. Intuitively, the more swaps are necessary, the more robust a particular election is.
Definition 4.
Let be a multiwinner rule. In the Robustness Radius problem we are given an election , a committee size , and an integer . We ask if it is possible to obtain an election by making at most swaps of adjacent candidates within the rankings in so that .
This problem is strongly connected to some other problems studied in the literature. Specifically, in the Destructive Swap Bribery problem [12, 24, 15] (DSB for short) we ask if it is possible to preclude a particular candidate from winning by making a given number of swaps. DSB was already used to study robustness of singlewinner election rules by Shiryaev et al. [24]. We decided to give our problem a different name, and not to refer to it as a multiwinner variant of DSB, because we feel that in the latter the goal should be to preclude a given candidate from being a member of any of the winning committees, instead of changing the outcome in any arbitrary way. In this sense, our problem is very similar to the Margin of Victory problem [19, 6, 25, 4], which has the same goal, but instead of counting single swaps, counts how many votes are changed.
We find that Robustness Radius tends to be computationally challenging. Indeed, we find polynomialtime algorithms only for the following simple rules.
Proposition 7.
SNTV Robustness Radius is polynomialtime solvable.
Proof.
Without loss of generality, the winning committee in the given election is . We begin by guessing which candidate which is not a committee member in will become a committee member in (where is the modified election, after performing at most swaps of consecutive candidates); let us denote this candidate by . Next we guess the Plurality score of in ; let us denote this score by .
There are two cases to consider now depending on whether the score of one of the current committee members is at most or not. If this is the case, then we proceed as follows. Intuitively, we do not need to decrease the score of any current committee member, thus we can concentrate on increasing the score of . To this end, we can proceed greedily, where in each greedy iteration we identify a voter which ranks in the highest position among all voters in the election, but not in the top position, and swapping towards the top position in this voter. If after iterations becomes a committee member, then we accept; otherwise, we reject.
The second case we shall consider is when the current score (that is, in ) of all the current committee members is higher than . If this is the case, then we proceed as follows. First we guess a committee member, say , whose score we will try to decrease. The point now is that our swaps shall increase the score of while decreasing the score of . To this end, we first guess the number of voters which currently (that is, in ) rank on top and will rank on top in . Then, we identify such voters which currently rank on top and, among these, rank the highest; we swap to the top in all of these votes. Similarly, we guess the number of votes which currently (that is, in ) do not rank nor on top and will rank on top in . Then, we identify such voters which currently do not rank nor on top and, among these, rank the highest; we swap to the top in all of these votes. Finally, we guess the number of voters which rank on top and will rank neither nor on top in . Then, we identify such voters and swap in those votes. If we have the budget to perform all the swaps described above and as a result enters a winning committee, then we accept; otherwise, we reject. ∎
Proposition 8.
Bloc Robustness Radius is polynomialtime solvable.
Proof.
The proof resembles the proof of the Proposition 7. The main observation is that it is sufficient to guess the candidate which is to become a committee member, and possibly also the committee member which is to cease being a committee member. Then, the task remains to increase the score of while increasing the score of . To this end, we consider the following input types of voters in :

Voters which give a point neither to nor to .

Voters which do not give a point to but give a point to .

Voters which give a point to but not to .

Voters which give a point to both and .
Then, we guess how voters of these input types will change in the resulting election , namely:

We guess the number of voters of input type (1) which will give a point to in .

We guess the number of voters of input type (2) which will (a) give a point to both and in ; (b) give a point to but not to in ; (c) will give a point neither to nor to in .

We do not consider voters of input type (3).

We guess the number of voters of input type (4) which will not give a point to in but will continue giving a point to .
After this guessing, we consider each of these types separately and greedily swap up and down to realize the guessing in the cheapest way possible. If we have the budget to do so and as a result enters a winning committee, then we accept; otherwise, we reject.
∎
Proposition 9.
Borda Robustness Radius is in .
Proof.
We follow similar lines as the proofs for SNTV (Proposition 7) and for Bloc (Proposition 8). Specifically, we guess the candidate which is to become a committee member, and possibly also the committee member which is to cease being a committee member. Then, the task remains to increase the score of while increasing the score of . Now we can guess the number of voters for which we will have a swap between and , and swap them first greedily (by starting from those for which it is the cheapest to do so); then, we guess the amount by which the score of needs to decrease and greedily realize it, and similarly guess the amount by which the score of needs to increase and greedily realize it too. If we have the budget to make it happen and as a result enters a winning committee, then we accept; otherwise, we reject. ∎
The rules in Propositions 7, 8, 9, are all robust, but not all robust rules have efficient Robustness Radius algorithms. In particular, a simple modification of a proof of Kaczmarczyk and Faliszewski [15, Theorem 3] shows that for Copeland rules (which are robust) we obtain hardness. We also obtain a general hardness for all Gehrlein weaklystable rules.
Corollary 2.
Copeland Robustness Radius is hard.
Theorem 10.
Robustness Radius is hard for each Gehrlein weaklystable rule.
Proof.
We reduce from the NPhard Set Cover problem, which, given sets over elements , asks for sets which, together, cover all elements. We create an election with the set of candidates , where:

,

,

,

.
We state the electorate in terms of the majority graph, where:

we make to be a cycle by adding an arc for each as well as the arc ,

we make to be a cycle by adding an arc for each as well as the arc ,

we make to be a cycle by adding an arc for each as well as the arc ,

we make to be a cycle by adding an arc for each as well as the arc ,

we let each () win over specifically by adding, for each set , one voter with the preference order .
We set the budget to be (we need to swap to the front of voters, corresponding to a set cover, and in each such voter, if we reduce from , then we need to swap it times).
As a result, we have that is the only Gehrlein weaklystable committee, thus any Gehrlein weaklystable rule shall select it. Initially, is not a Gehrlein weaklystable rule, but by choosing a set cover and swapping to the first position in the corresponding voters, it becomes a Gehrlein weaklystable committee. ∎
Without much surprise, we find that Robustness Radius is also hard for STV and for . For these rules, however, the hardness results are, in fact, significantly stronger. In both cases it is hard to decide if the election outcome changes after a single swap, and for STV the result holds even for committees of size one ( with committees of size one is simply the singlewinner Borda rule, for which the problem is polynomialtime solvable [24]).
Theorem 11.
CC Robustness Radius is hard and hard with respect to the size of the committee even if the robustness radius is one.
Proof.
We reduce from the Regular Multicolored Independent Set problem which asks, given an undirected graph where each vertex has one of colors and exactly neighbors, whether there is an colored independent set that is, a size set of pairwise nonadjacent vertices containing one vertex from each color class. This problem is hard for the parameter [10][Corollary 13.8].
In order to show that CC Robustness Radius is hard we will provide a parameterized reduction with respect to the parameters , and the committee size (for the definition of the parameterized reductions we refer the reader to the book of Cygan et al. [10]).
Given an instance of Regular Multicolored Independent Set we construct a CC Robustness Radius instance as follows. To this end, let and .
Candidates, committee size, and construction idea. The set of candidates consists of the vertex set of the graph , the set of special candidates, the set of safe candidates, and the set of dummy candidates (its size is polynomial in and will become clear later). We set the committee size .
The idea is as follows. The election will have at least the safe committee as (possibly unique) winning committee. If there is an colored independent set, then also every committee with being an colored independent set wins and one can easily reduce the set of winning committees by a single swap. However, if there is no colored independent set, then the safe winning committee is unique and the total dissatisfaction (recall that we define the concept of dissatisfaction at the beginning of the appendix) of the next best committee differs by at least four (thus, a single swap cannot change the set of winning committees).
Dummy candidates. In this construction, we will introduce a lot of dummy candidates which simplify the construction and the respective proof significantly. The idea behind is as follows: Our construction ensures that the total dissatisfaction of all voters given some optimal winning committee is upperbounded by . Now in our construction, every dummy candidate appears at most in one vote at some of the first positions and in all other voter in later positions. This ensures that a dummy candidate can never be part of a winning committee. To further simplify the writeup we use the symbol “” to denote a block of many consecutive dummy candidates. It is easy to verify that there is no need to specify the concrete ordering of the candidates behind such a block of many dummy candidates.
Voter groups. In the following, we describe the voters of our election in four groups, each playing a specific role in the construction. We briefly mention the respective roles and formally prove them later.
 Special candidates group.

This group consists of copies of the vote “” and ensures that one must include candidate in the committee.
 Safe committee group.

For each color we add copies of the vote “”. This group is essential to ensure that the safe winning committee is indeed among the winning committees.
 Vertex selection group.

For each color we add the following votes where each vertex appears exactly once in each of the first positions. Technically, we start with the following votes
and replace each vertex that is not of color by a new dummy candidate. The intuition behind this group is to ensure that every further winning committee (not being the safe winning committee) must select exactly one vertex of each color.
 Independent set detector group.

For every edge we introduce two pairs of votes as follows:
The intuition of this group is to ensure that one cannot put two adjacent vertices into a winning committee.
This completes the construction which can be performed in polynomial time.
Possible committees. Before we formally prove the correctness of our construction, we discuss several important facts about possible winning committees for the constructed election.
First, observe that the safe committee causes total dissatisfaction : There is no dissatisfaction from the special candidates group and from the safe committee group, there are dissatisfaction points from the independent set detector group and dissatisfaction points from the vertex selection group.
Second, observe that also every committee with being an colored independent set causes total dissatisfaction : There is no dissatisfaction from the special candidates group, dissatisfaction points from the safe committee group, dissatisfaction points from the vertex selector group, and dissatisfaction points from the independent set detector group. These numbers are obvious for the former two groups, so let us focus on the latter two groups. As for the safe committee group, note that any committee that contains exactly one vertex of each color causes dissatisfaction points from this group: For every color , the corresponding committee member appears exactly once in every position for some of of the voters corresponding to color . There is one voter with dissatisfaction (where the committee member is at the top position), one voter with dissatisfaction (where the committee member is at the second position), and so on. Thus the total dissatisfaction is . As for the independent set detector group, let us distinguish between (1) edges which are not covered by the committee, that is, neither nor are part of the committee, (2) edges which are covered once by the committee, that is, either or is part of the committee, and (3) edges which are covered twice by the committee, that is, both and are part of the committee. Observe that the committee causes dissatisfaction points for every four voters corresponding to an uncovered edge, dissatisfaction points for every four voters corresponding to an edge that is covered once, and no dissatisfaction points for every four voters corresponding to an edge that is covered twice. Now, since corresponds to an independent set of size and every vertex has neighbors, it holds that there are dissatisfaction points from the independent set detector group.
Next, we show that any other committee causes total dissatisfaction at least . To this end, we distinguish between five cases for possible committees and call a committee that causes total dissatisfaction at most almost winning.
Every committee causes total dissatisfaction at least . When is not part of the committee, then up to voters from the special candidates group receive dissatisfaction at least one (in best case, they are represented by their secondbest choice) while the rd voter receives dissatisfaction at least . Summarizing, must be part of every (almost)winning committee.
Every committee causes total dissatisfaction at least . Consider the dissatisfaction of the voters when they would only be represented by . So far, there is no dissatisfaction from the special candidates group, dissatisfaction points from the safe committee group, dissatisfaction points from the vertex selector group, and dissatisfaction points from the independent set detector group. Hence, the remaining candidates must further reduce the total dissatisfaction by . However, no single candidate can reduce the total dissatisfaction by more than . This is obvious for safe candidates. For vertex candidates, observe that the dissatisfaction in the vertex selection group will be reduced by at most —and only if the committee does not already contain a vertex of the same color—while the dissatisfaction in the independent set detector group will be reduced by at most —and only if none of the neighbors is already part of the committee. Summarizing, an (almost) winning committee cannot contain both, candidate and candidate .
Every committee that does not contain all candidates from causes total dissatisfaction (much) larger than . Without loss of generality assume that . Then at least copies of the vote “” are dissatisfied by at least . Every committee that does not contain one vertex of each color causes total dissatisfaction (much) larger than . Without loss of generality assume that does not contain a vertex of color . Then, of the votes corresponding to color are dissatisfied by at least . By very similar arguments one can show that every committee causes total dissatisfaction (much) larger than . Summarizing, an (almost)winning committee that is not the safe committee must be of the form with containing one vertex of each color .
Finally, every committee with not being an colored independent set causes total dissatisfaction at least . There is no dissatisfaction from the special candidates group, dissatisfaction points from the safe committee group, dissatisfaction points from the vertex selector group, and at least dissatisfaction points from the independent set detector group. For the latter group, let be the number of edges between vertices from . Thus, there are edges that are covered twice, edges that are covered once, and all remaining edges are uncovered. The total dissatisfaction of the corresponding voters is at least . Since is not an independent set, we have and the claim follows.
Correctness. The correctness easily follows from the above discussion: On the one hand, if graph does not contain an colored independent set, then the safe committee is the only winning committee with total dissatisfaction and every other committee has dissatisfaction at least . Thus, a single swap cannot change the set of winning committees. On the other hand, if graph does contain an colored independent set, then the safe committee is a nonunique winning committee. It is easy to verify that the safe committee does not win anymore if one swaps candidate with some candidate in some vote from the safe committee group. ∎
Theorem 12.
STV Robustness Radius is hard even for and .
Proof.
We give a reduction from STV Winner Determination—the problem of deciding whether for a given election, a given candidate can become an STV winner. This problem is known to be hard [9][Theorem 4] for . Let be an instance of the STV Winner Determination problem. In we are given an election , , and a distinguished candidate ; we ask if there exists a valid run of STV such that becomes a winner in . W.l.o.g., we can assume that is ranked first by some voter.
From we construct an instance of the STV Robustness Radius problem as follows. We fix the new set of candidates to ; here, is a dummy candidate needed by our construction. For each , we put in ’s preference ranking right behind , and add two copies of such modified vote to , we call such votes nondummy. Additionally, we add dummy voters who rank first, second and all the remaining candidates next (i some fixed arbitrary order). It is easy to check that is the unique winner in such instance. Next, observe that if we want to change the outcome of an election with a single swap, then we definitely need to swap and for one of the voters who rank first. Let us consider such modified election, call it .
Observe that if is a possible winner in , then is also a possible winner in . Indeed, STV can first eliminate all candidates except for and . In such truncated profile, there will be voters who prefer to and voters who prefer to ; hence will become a winner.
If is not a possible winner in , then will be eliminated before any other candidate from in
. Indeed, in each sequence of eliminations performed by STV, there will be a moment when there are still some remaining candidates in
and that each such candidate is ranked first by at least 2 more nondummy voters than ; as a result each such candidate will be ranked first by more (dummy and nondummy) voters than . In particular, will be removed from the election before some candidate from , and so, also before . After is removed from , there will be at least voters who rank first, thus this candidate will become the unique winner. Consequently, we have shown that an outcome of an election can change with a single swap if and only if the answer to the original instance is “yes”. This completes the proof. ∎For the case of , the proof of Theorem 11 gives much more than stated in the theorem. Indeed, our construction shows that the problem remains hard even if we are given a current winning committee as part of the input. Further, the same construction gives the following corollary (whose first part is sometimes taken for granted in the literature, but has not been shown formally yet).
Corollary 3.
The problem of deciding if a given candidate belongs to some winning committee (for a given election and committee size) is both hard and hard.
We conclude this section by providing algorithms for Robustness Radius. An algorithm for a given parameter (e.g., the number of candidates or the number of voters) must have running time of the form , where is the value of the parameter and
is the length of the encoding of the input instance. Using the standard approach of formulating integer linear programs and invoking the algorithm of Lenstra
[17], we find that Robustness Radius is in when parametrized by the number of candidates.Proposition 13.
Robustness Radius for Copeland, NED, STV, and is in (parametrized by the number of candidates).
For STV and we also get algorithms parametrized by the number of voters. For the case of STV, we assume that the value of is such that we never need to “delete nonexistent voters” and we refer to committee sizes where such deleting is not necessary as normal. For example, a committee size is not normal if (where is the number of voters). Another example is to take and : We need to delete voters for each committee member, which requires deleting voters out of .
Theorem 14.
STV Robustness Radius is in when parametrized by the number of voters (for normal committee sizes).
Proof.
Let be our input election and let be the size of the desired committee. For each candidate , “rank of ” is defined as .
First, we prove that a candidate with a rank higher than cannot be a member of a winning committee. Let us assume towards a contradiction that there exists a candidate with who is a member of some winning committee . When STV adds some candidate to the committee (this happens when the number of voters who rank such a candidate first exceeds the quota ), it removes this candidate and at least one voter from the election. Thus, before were included in , STV must have removed some candidate from the election without adding it to (this is so because had to be ranked first by some voter to be included in the committee; for to be ranked first, STV had to delete at least candidates, so by the assumption that the committee size is normal, not all of them could have been included in the committee). Since STV always removes a candidate with the lowest Plurality score, and at the moment when was removed the Plurality score of was equal to zero, we have that the Plurality score of also must have been zero. Consequently, removing from the election did not affect the top preferences of the voters, and STV, right after removing , removed another candidate with zero Plurality score from the election. By repeating this argument sufficiently many times, we conclude that must have been eventually eliminated, and so could not have been added to . This gives a contradiction and proves our claim.
The same reasoning also shows that the number of committees winning according to STV is bounded by a function of : In each step either one of at most voters is removed, or all candidates who are not ranked first by any voter are removed from the election (which leaves at most candidates in the election).
Second, we observe that the robustness radius for our election is at most . Indeed, we can take any winning candidate, and with at most swaps we can push him or her to have rank at least . Such a candidate will no longer be a member of and, so, the outcome of the election will change.
Third, we observe that in order to change the outcome of an election, we should only swap such candidates that at least one of them has rank at most . Indeed, consider a candidate with . After swaps, the rank of this candidate would still be above , so he or she still would not belong to any winning committee. Thus, a swap of any two candidates with ranks higher than does not belong to any of the sequences of at most swaps that change the election result (the exact positions of these two candidates would have no influence on the STV outcome).
As a result, it suffices to focus on the candidates with ranks at most . There are at most such candidates. Consequently, there are at most possible long sequences of swaps which we need to check in order to find the optimal one. This completes the proof. ∎
The algorithm for the case of is more involved. Briefly put, it relies on finding in time (with respect to the number of voters) either the unique winning committee or two committees tied for victory. In the former case, it combines bruteforce search with dynamic programming, and in the latter either a single swap or a clever greedy algorithm suffice. Due to clarity, we start with presenting the first phase, i.e., finding the unique winning committee or two tied committees, as a separate proposition.
Proposition 15.
There is an algorithm that runs in time with respect to the number of voters and, given an election and a committee size , checks whether the election has a unique CC winning committee (in which case it outputs this committee) or whether there is more than one CC winning committee (in which case it outputs some two winning committees).
Proof.
Let be an input election and let be the committee size. We let be the number of voters. If , then every winning committee consists of each voter’s most preferred candidate and sufficiently many other candidates to form a committee of size exactly . In this case the algorithm can easily provide the required output, so we assume that . (To avoid triviality, w.l.o.g., we also assume that there are more than candidates.)
Our algorithm proceeds by considering all partitions of into disjoint sets (there are at most such partitions). For a partition the algorithm proceeds as follows (intuitively, the voters in each group are to be represented by the Borda winner of the election ):

For each election we compute the set of candidates that are Borda winners of .

If each is a singleton and all ’s are distinct, then we store a single committee . Otherwise, it is possible to form two distinct committees, and , such that for each , and ;^{4}^{4}4We can form and as follows. First, for each we include the lexicographically first member of in and the lexicographically last one in . If at least one contains more than one candidate, then we will obtain distinct committees, possibly of size smaller than . If each is a singleton, then we obtain two identical committees that certainly contain less than candidates (otherwise we would not enter this part of the algorithm). In either case, since there are more than candidates in the election, it is possible to supplement and with candidates so that they are distinct and of size exactly . we store both and .
We check if among the stored committees there is a unique committee such that every other stored committee has lower CC score. If such a committee exists then we output it as the unique winning committee. Otherwise, there are two stored committees, and , that both have CC score greater than or equal to that of every other stored committee. We output and as two committees tied for winning (if there is more than one choice for and then we pick one pair arbitrarily). ∎
Before we move on to the proof of the next theorem, we need to introduce some additional notation. Let be some election and let be some voter in . By we mean the candidate ranked first by . By we mean the set , that is, the set of candidates that are ever ranked first in election . For a committee , the representative of a voter is the member of that ranks highest. Finally, for committee and voter , we define to be the position of ’s representative from in ’s vote.
Theorem 16.
CC Robustness Radius is in when parameterized by the number of voters.
Proof.
Let be our input election and let be the committee size. We let be the number of candidates. Using Proposition 15, we check whether there is a unique CC winning committee in . Depending on the result, we proceed in one of the following two ways.
There is a unique winning committee . We first describe a function that encapsulates the effect of shifting forward a particular candidate within a given set of votes. For each voter , each candidate , and each nonnegative integer , we define to be the vote obtained from that of by shifting by positions forward, and we define:
In other words, is the difference between the Borda scores of and the highestranked member of in vote with shifting positions forward.
Let be some subset of voters, and let us rename the voters so that . For each candidate and each nonnegative integer , we define:
Intuitively, specifies how many points more would receive from the voters in as their representative than these voters would assign to their representatives from , if we shifted by positions forward in an optimal way.
We assume that for each choice of and . We can compute in polynomial time using dynamic programming and the following formula (for each ):^{5}^{5}5In fact, it is possible to compute using a greedy algorithm, but the dynamic programming formulation is far easier and allows us to sidestep many special cases, such as what happens if is him or herself a member of .
With the function in hand, we are ready to describe the algorithm. We consider every partition of into disjoint subsets ; let us fix one such partition. Our goal is to compute the smallest nonnegative integer such that there is a sequence of nonnegative integers that adds up to , and a sequence of (not necessarily distinct) candidates so that:
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