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If [MATH] is a positive literal in [MATH] the edges [MATH] [MATH] and [MATH] are added. If [MATH] is a negative literal in [MATH] the edges [MATH] [MATH] and [MATH] are added. If [MATH] does not occur in [MATH] the edges [MATH] [MATH] [MATH] and [MATH] are added, see Figure 10 for an example. In the following they argu... |
We now modify the variable gadget as shown in Figure 9(c) so that at least one of the nodes [MATH] must be in every resolving set. Note that whenever one of the nodes [MATH] is in a resolving set it can also be substituted by one of the nodes [MATH] |
[MATH] Conclusion We have shown that Metric Dimension can be solved in polynomial time on graphs having a minimum resolving set with a bounded number of resolving vertices in every EBC. Even more the algorithm even can compute a minimum resolving set in polynomial time under these restrictions. However, the problem rem... |
# Source: arxiv 1806.10460 # Title: On Coalitional Manipulation for Multiwinner Elections: Shortlisting # Sections: all # Downloaded: 2026-03-03T01:44:41.318767+00:00 |
On Coalitional Manipulation for Multiwinner Elections: Shortlisting Abstract Shortlisting of candidates—selecting a group of “best” candidates—is a special case of multiwinner elections. We provide the first in-depth study of the computational complexity of strategic voting for shortlisting based on the perhaps most ba... |
Keywords: computational social choice; utility aggregation; strategic voting; parameterized computational complexity; tie-breaking; SNTV Bloc |
Introduction A university wants to select the two favorite pieces in classical style to be played during the next graduation ceremony. The students were asked to submit their favorite pieces. Then a jury consisting of seven members (three juniors and four seniors) from the university staff selects from the six most fre... |
The senior jury members meet every Friday evening and discuss important academic issues including the graduation ceremony music selection processes, why “movie background noise” recently counts as classical music, and the influence of video games on the ability of making important decisions. During such a meeting they ... |
Already this toy example above (which will be the basis of our running example throughout the paper) illustrates important aspects of strategic voting in multiwinner elections. In case of coalitional manipulation for single-winner elections (where a coalition of voters casts untruthful votes in order to influence the o... |
Multiwinner voting rules come up very naturally whenever one has to select from a large set of candidates a smaller set of “the best” candidates. Surprisingly, although at least as practically relevant as single-winner voting rules, the multiwinner literature is much less developed than the single-winner literature. In... |
short list Shortlisting comes very naturally in the context of selection committees; for instance, for human resources departments that need to select, for a fixed number of positions, the best qualified applicants. A standard way of candidate selection in the context of shortlisting is to use scoring-based voting rule... |
To come up with a useful framework for coalitional manipulation for multiwinner elections, we first have to identify the exact mathematical model and questions to be asked. A couple of straightforward extensions of coalitional manipulation for single-winner elections or (non-coalitional) manipulation for multiwinner el... |
We address the aforementioned issue of modeling coalitional manipulation for multiwinner election by extending a single-manipulator model for multiwinner rules of Meir et al. ( 2008 . In their work, the manipulator specifies the utility of each candidate and the utility for a candidate group is obtained by adding up th... |
Our Contributions. We devise a formal description of coalitional manipulation in multiwinner elections arriving at a new, nontrivial model capturing two types of manipulators’ attitudes and a few natural ways of utility aggregation. To this end, in our model, we distinguish between optimistic and pessimistic manipulato... |
Using our model, we analyzed the computational complexity of finding a successful manipulation for a coalition of voters, assuming elections under rules from the family of [MATH] Bloc voting rules. We show that, even for these fairly simple rules, the computational complexity of coalitional manipulation is diverse. In ... |
Related Work. To the best of our knowledge, there is no previous work on coalitional manipulation in the context of multiwinner elections. We refer to recent textbooks for an overview of the huge literature on single-winner (coalitional) manipulation (Rothe, 2015 ; Brandt et al., 2016 . Most relevant to our work, |
Lin ( 2011 showed that coalitional manipulation in single-winner elections under [MATH] -Approval is solvable in linear time by a greedy algorithm. |
Meir et al. ( 2008 introduced (non-coalitional) manipulation for multiwinner elections. While pinpointing manipulation for several voting rules as NP-hard, they showed that manipulation remains polynomial-time solvable for |
Bloc (which can be interpreted as a multiwinner equivalent of [MATH] -Approval). Obraztsova et al. ( 2013 extended the latter result for different tie-breaking strategies and identified further tractable special cases of multiwinner scoring rules but conjectured manipulation to be hard in general for (other) scoring ru... |
coalitional manipulation. Organization. Section 2 introduces basic notation and formal concepts. In Section 3 , we develop our model for coalitional manipulation in multiwinner elections. Its variants respect different ways of evaluating candidate groups (utilitarian vs. egalitarian) and two kinds of manipulators behav... |
Preliminaries For a positive integer [MATH] , let [MATH] . We use the toolbox of parameterized complexity (Cygan et al., 2015 ; Downey and Fellows, 2013 ; Flum and Grohe, 2006 ; Niedermeier, 2006 to analyze the computational complexity of our problems in a fine-grained way. To this end, we always identify a parameter |
[MATH] that is typically a positive integer. We call a problem parameterized by [MATH] fixed-parameter tractable (in [MATH] ) if it is solvable in [MATH] time, where [MATH] is the size of a given instance encoding, [MATH] |
is the value of the parameter, and [MATH] is an arbitrary computable (typically super-polynomial) function. To preclude fixed-parameter tractability, we use an established complexity hierarchy of classes of parameterized problems, [MATH] . It is widely believed that all inclusions are proper. The notions of hardness fo... |
[MATH] which is a more explicit way of expressing a parameter [MATH] An election [MATH] consists of a set [MATH] of [MATH] candidates and a multiset [MATH] of [MATH] votes. Votes are linear orders over [MATH] —for example, for |
[MATH] we write [MATH] to express that candidate [MATH] is the most preferred and candidate [MATH] is the least preferred according to vote [MATH] . We write [MATH] if the corresponding vote is clear from the context. |
multiwinner voting rule is a function that, given an election [MATH] and an integer [MATH] , outputs a family of co-winning size- [MATH] subsets of [MATH] representing the co-winning [MATH] -excellence-groups . We use [MATH] -egroup as an abbreviation for [MATH] -excellence-group; we also use egroup if the size of an e... |
We consider scoring rules —multiwinner voting rules that assign points to candidates based on their positions in the votes. By [MATH] , we denote the total number of points that candidate [MATH] obtains, and we use [MATH] when restricting the election to a subset [MATH] of voters. A (multiwinner) scoring rule selects a... |
Example 1 Referring back to our introductory example, we have a set [MATH] of candidates and a set [MATH] of voters. The voters [MATH] [MATH] , and [MATH] represent the three junior jury members, whereas [MATH] [MATH] and [MATH] [MATH] represent, respectively, the Beethoven and Mozart enthusiasts among the senior jury ... |
[EQUATION] Following the introductory example, we are choosing an egroup of size [MATH] Using the Bloc multiwinner voting rules (which coincides with our introductory example), the winning [MATH] -egroup consist of candidates |
[MATH] and [MATH] . However, under the SNTV voting rule the situation would change, and the winners would be [MATH] and [MATH] SNTV and Bloc alike output a single winning egroup in this example, and thus tie-breaking is ineffective. |
To select a single [MATH] -egroup from the set of co-winning [MATH] -egroups one has to consider tie-breaking rules. multiwinner tie-breaking rule is a mapping that, given an election and a family of co-winning [MATH] -egroups, outputs a single [MATH] -egroup. Among them, there is a set of natural rules that is of part... |
is described not only by the preference order [MATH] of the candidates but also by a utility function [MATH] To cover this in the tie-breaking process, coalition-specific |
tie-breaking rules get—in addition to the original election, the manipulators’ votes, and the co-winning excellence-groups—the manipulators’ utility functions in the input. The formal implementations of these rules and their properties are discussed in Subsection 3.2 |
Model for Coalitional Manipulation In this section, we formally define and explain our model and the respective variants. To this end, we discuss how we evaluate a [MATH] -egroup in terms of utility for a coalition of manipulators and introduce tie-breaking rules that model optimistic or pessimistic viewpoints of the m... |
3.1 Evaluating [MATH] -egroups As already discussed in the introduction, one should not extend the model of coalitional manipulation for single-winner elections to multiwinner elections in the simplest way (e.g., by assuming that the manipulators agree on some distinguished candidate or on some distinguished egroup). I... |
Meir et al. ( 2008 and assume that we are given a utility function over the candidates for each manipulator and a utility level which, if achieved, indicates a successful manipulation. Meir et al. ( 2008 compute the utility of an egroup by summing up the utility values the manipulator assigns to each member of the egro... |
At first glance, summing up the utility values assigned by each manipulator to each member of an egroup seems to be the most natural extension for a coalition of manipulators. However, this utilitarian variant does not guarantee single manipulators to gain non-zero utility. In extreme cases it could even happen that so... |
Example 2 Example 2 Consider the election [MATH] where [MATH] is a set of candidates and [MATH] is the following multiset of three votes: |
[EQUATION] Additionally, consider two manipulators, [MATH] and [MATH] , that report utilities to the candidates as depicted in the table below. |
[MATH] [MATH] Let us analyze the winning [MATH] -egroup under the SNTV voting rule. Observe that if the manipulators vote sincerely, then together they give one point to [MATH] and one to [MATH] (one point from each manipulator). Combining the manipulators’ votes with the non-manipulative ones, the winning |
[MATH] -egroup consists of candidates [MATH] and [MATH] that both have score two; no other candidate has greater or equal score, so tie-breaking is unnecessary. The value of such a group is equal to seven according to the utilitarian evaluation variant. Manipulator [MATH] ’s utility is seven. However, both manipulators... |
[MATH] (according to the utilitarian variant). Observe that in spite of growth of the total utility, the utility value gained by [MATH] , which is one, is lower than in the case of sincere voting. |
In Example 2 manipulator [MATH] devotes its satisfaction to the utilitarian satisfaction of the group of the manipulators; that is, [MATH] is worse off voting strategically compared to voting sincerely. Despite this issue, however, the utilitarian viewpoint can be justified if the manipulators are able to compensate su... |
We formalize the described variants of [MATH] -egroup evaluation (for [MATH] manipulators) with Definition 1 Definition 1 Given a set of candidates [MATH] , a [MATH] -egroup [MATH] [MATH] , and a family of manipulators’ utility functions [MATH] where [MATH] , we consider the following functions: |
[MATH] [MATH] [MATH] Intuitively, these functions determine the utility of a [MATH] -egroup [MATH] according to, respectively, the utilitarian and the egalitarian variants of evaluating [MATH] by a group of [MATH] manipulators (identifying manipulators with their utility functions). We omit subscript [MATH] when [MATH]... |
Example 3 Example 3 Consider our example set of candidates [MATH] and two manipulators [MATH] [MATH] whose utility functions over the candidates are depicted in the table below. |
[MATH] [MATH] Then, evaluating the utility of [MATH] -egroup [MATH] applying the three different evaluation variants gives: [MATH] |
[MATH] [MATH] Analyzing Example 3 , we observe that we can compute the utilitarian value of egroup [MATH] by summing up the overall utilities that each candidate in [MATH] contributes to all manipulators; for instance candidate [MATH] always contributes the utility of [MATH] to the manipulators, independently of other ... |
Observation 1 Without loss of generality, one can assume that there is a single non-zero valued utility function over the candidates under the utilitarian or candidate-wise egalitarian evaluation. |
Proof. Consider a multiset of manipulators’ utility functions [MATH] and a [MATH] -egroup [MATH] For the utilitarian variant, create a new utility function [MATH] that assigns to each candidate the sum of utilities given to this candidate by all manipulators; that is, [MATH] for all [MATH] Technically, we need also a s... |
that assigns to each candidate the utility of zero. We construct a new family [MATH] consisting of function [MATH] and [MATH] copies of function [MATH] . Since function [MATH] is the only one which gives non-zero utility in family [MATH] and for each candidate this function returns the sum of utilities given to a candi... |
[MATH] We follow a similar strategy proving Observation 1 for the candidate-wise variant. We introduce a function [MATH] defined as [MATH] for each candidate [MATH] . Then we create a new family of utility functions [MATH] with function [MATH] and [MATH] zero-valued function [MATH] . Naturally, [MATH] because the value... |
3.2 Breaking Ties Before formally defining our tie-breaking rules, we briefly discuss some necessary notation and central concepts. Consider an election [MATH] , a size [MATH] for the egroup to be chosen, and a scoring-based multiwinner voting rule [MATH] . We can partition the set of candidates [MATH] |
into three sets [MATH] [MATH] , and [MATH] as follows: The set [MATH] contains the confirmed candidates , that is, candidates that are in all co-winning [MATH] -egroups. The set [MATH] contains the pending candidates , that is, candidates that are only in some co-winning [MATH] -egroups. The set [MATH] contains the rej... |
[MATH] , and that every candidate from [MATH] receives fewer points than every candidate from [MATH] . Additionally, all candidates in [MATH] receive the same number of points. |
We define the following families of tie-breaking rules which are considered in this work. In order to define optimistic and pessimistic rules, we assume that in addition to [MATH] [MATH] , and [MATH] , we are given a family of utility functions which are used to evaluate the [MATH] -egroups as discussed in |
Subsection 3.1 Lexicographic [MATH] A tie-breaking [MATH] belongs to [MATH] if and only if ties are broken lexicographically with respect to some predefined order [MATH] of the candidates from [MATH] . That is, [MATH] selects all candidates from [MATH] and the top [MATH] candidates from [MATH] with respect to [MATH] |
Optimistic [MATH] [MATH] A tie-breaking belongs to [MATH] if and only if it always selects some [MATH] -egroup [MATH] such that [MATH] and there is no other [MATH] -egroup [MATH] |
with [MATH] and [MATH] Pessimistic [MATH] [MATH] A tie-breaking belongs to [MATH] if and only if it always selects some [MATH] -egroup [MATH] such that [MATH] and there is no other [MATH] -egroup [MATH] |
with [MATH] and [MATH] We remark that the definitions above come in two, substantially different variants. For each lexicographic tie-breaking rule, there is always exactly one egroup that will be selected by the rule for a particular set of pending set candidates. However, it is not the case for the families of pessim... |
3.3 Limits of Lexicographic Tie-Breaking From the above discussion, we can conclude that lexicographic tie-breaking is straightforward in the case of scoring-based multiwinner voting rules. Basically any subset of the desired cardinality from the set of pending candidates can be chosen. In particular, the best pending ... |
It remains to be clarified whether one can find a reasonable order of the pending candidates in order to model optimistic or pessimistic tie-breaking rules in a simple way. We show that this is possible for every |
[MATH] [MATH] [MATH] , using the fact that in these cases we can safely assume that there is only one non-zero valued utility function (see |
Observation 1 ). On the contrary, there is a counterexample for [MATH] and [MATH] . On the way to prove these claims we need to formally define what it means that one family of tie-breaking rules can be used to simulate another family of tie-breaking rules. |
Definition 2 We call a triplet consisting of an election with candidate set [MATH] , a size of an egroup, and a family of utility functions a tie-breaking perspective over [MATH] . Let [MATH] [MATH] |
[MATH] , and [MATH] be all possible tie-breaking perspectives over [MATH] admitting, respectively, a set [MATH] of confirmed candidates, a set [MATH] of pending candidates, a family [MATH] of utility functions, and a size [MATH] of an egroup. For some non-empty subset [MATH] of the set [MATH] let [MATH] |
Then, for two families [MATH] and [MATH] of tie-breaking rules we say that [MATH] can [MATH] -simulate [MATH] if there exists a rule [MATH] such that for each tie-breaking perspective in [MATH] there exists a rule [MATH] such that |
[MATH] and [MATH] yield the same output for this perspective. We call rule [MATH] [MATH] -simulator At first glance, Definition 2 might seem overcomplicated. However, it is tailored to grasp different degrees of simulation possibilities. On the one hand, one can always find a lexicographic order and use it for breaking... |
Observation 2 Let [MATH] be a fixed set of candidates, [MATH] be a set of confirmed candidates, [MATH] be a set of pending candidates, and [MATH] be a size of an egroup. Let [MATH] and [MATH] . The family of lexicographic tie-breaking rules does not |
[MATH] -simulate [MATH] Proof. Suppose [MATH] [MATH] , and [MATH] that is, we are going to select either [MATH] or [MATH] who are tied. Let us fix a family [MATH] of utility functions such that [MATH] and |
[MATH] For the family [MATH] of utility functions clearly [MATH] selects candidate [MATH] Now, consider a family [MATH] of utility functions where [MATH] assigns utility one to candidate [MATH] and zero otherwise. For this family, [MATH] selects candidate [MATH] . This means that we cannot find a [MATH] -simulator [MAT... |
of tie-breaking rules because in the first case [MATH] would have to choose [MATH] and in the second case [MATH] would have to be chosen. This is impossible using a single preference order over [MATH] Similar families of functions (obtained by exchanging each one with zero and vice versa) yield a proof for [MATH] as we... |
Next, we show that for some cases it is sufficient to fix just the utility functions in order to simulate optimistic or pessimistic tie-breaking rules (see Proposition 1 ). For other cases, however, one has to fix all: confirmed candidates, pending candidates, utility functions, and the size of an egroup (see Propositi... |
Proposition 1 Let [MATH] be a set of candidates, [MATH] be a family of utility functions, [MATH] , and [MATH] . Let [MATH] and [MATH] . Then the family of lexicographic tie-breaking rules [MATH] can |
[MATH] -simulate [MATH] , and a [MATH] -simulator [MATH] can be found in [MATH] time. Proof. Recall from Observation 1 that if [MATH] then there is always a set of utility functions with just one non-zero valued utility function [MATH] that is equivalent to [MATH] Hence, we compute such a function [MATH] in [MATH] time... |
with some utility function [MATH] if [MATH] implies [MATH] for optimistic tie-breaking and [MATH] implies [MATH] for pessimistic tie-breaking. Any lexicographic tie-breaking rule defined by an order [MATH] that is consistent with the utility function [MATH] simulates |
[MATH] We compute a consistent order by sorting the candidates according to [MATH] in [MATH] time. Proposition 1 describes a strong feature of optimistic utilitarian and candidate-wise egalitarian tie-breaking and their pessimistic variants. Intuitively, the proposition says that for these tie-breaking mechanisms one c... |
Proposition 2 Let [MATH] be a set of candidates, [MATH] be a family of utility functions, [MATH] be a set of confirmed candidates, [MATH] be a set of pending candidates, and |
[MATH] be a size of an egroup. For each [MATH] [MATH] , the lexicographic tie-breaking family of rules does not [MATH] -simulate [MATH] assuming |
[MATH] Proof. From Observation 2 we already know that the family of lexicographic tie-breaking rules cannot [MATH] -simulate the family of egalitarian pessimistic tie-breaking rules or the family of egalitarian optimistic tie-breaking rules. |
Next, we build one counterexample for each of the remaining size-three subsets of [MATH] to show our theorem. To this end, let us fix a set of candidates [MATH] |
(compatible with our running example) and a family [MATH] of utility functions as depicted in the table below. [MATH] [MATH] [MATH] |
[MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] First, we prove that the family [MATH] cannot [MATH] -simulate [MATH] for [MATH] . Let us fix [MATH] [MATH] . We consider the optimistic variant of egalitarian tie-breaking for [MATH] , so we ar... |
Second, we prove that the family [MATH] cannot [MATH] -simulate [MATH] for [MATH] . This time, we fix [MATH] [MATH] . We construct the first case by setting [MATH] . Using the fact that in both functions candidate [MATH] has utility zero, we choose exactly the same candidate as in the proof of [MATH] -simulation for th... |
[MATH] ; that is, for the optimistic variant, the winning [MATH] -egroup is [MATH] and [MATH] . Consequently, [MATH] precedes [MATH] and [MATH] in the potential [MATH] -simulator’s lexicographic order. Towards a contradiction, we set [MATH] . The situation is exactly the same as in the proof of the [MATH] -simulation c... |
[MATH] -egroup consists of [MATH] and [MATH] which ends the proof for the optimistic case. By almost the same argument, the result holds for the pessimistic variant. |
Finally, we prove that the family [MATH] cannot [MATH] -simulate [MATH] for [MATH] . We fix [MATH] [MATH] . For the first case we pick |
[MATH] . The best egalitarian evaluation happens for the [MATH] -egroup consisting of [MATH] and [MATH] . This imposes that, in the potential [MATH] -simulator’s order, [MATH] and [MATH] precede the remaining candidates (in particular, [MATH] precedes [MATH] ). However, for |
[MATH] the best [MATH] -egroup changes to that consisting of [MATH] and [MATH] which gives a contradiction ( [MATH] precedes [MATH] ). As in the previous cases, the same argument provides a proof for the pessimistic variant. |
Proposition 2 implies that pessimistic and optimistic egalitarian tie-breaking cannot be simulated without having full knowledge about an election. In terms of computational complexity, however, finding winners for pessimistic egalitarian tie-breaking remains tractable whereas the same task for optimistic egalitarian t... |
Complexity of Tie-Breaking It is natural to ask whether the tie-breaking rules proposed in Subsection 3.2 are practical in terms of their computational complexity. If not, then there is little hope for coalitional manipulation because tie-breaking might be an inevitable subtask to be solved by the manipulators. Indeed,... |
Clearly, we can apply every lexicographic tie-breaking rule that is defined through some predefined order of the candidates in linear time. Hence, we focus on the rules that model optimistic or pessimistic manipulators. To this end, we analyze the following computational problem. |
[MATH] [MATH] Input: A set of candidates [MATH] partitioned into a set [MATH] of pending candidates and a set [MATH] of confirmed candidates, the size [MATH] of the excellence-group such that [MATH] , a family of manipulators’ utility functions [MATH] where [MATH] , and a non-negative, integral evaluation threshold [MA... |
Question: Is there a size- [MATH] set [MATH] such that [MATH] is selected according to [MATH] [MATH] , and [MATH] [MATH] -Tie-Breaking [MATH] -TB |
Naturally, we may assume that the number of candidates and the number of utility functions are polynomially bounded in the size of the input. However, both the evaluation threshold and the utility function values are encoded in binary. |
Note that an analogous problem has not been considered for single-winner elections. The reason behind this is that, for single-winner elections, optimistic and pessimistic tie-breaking rules can be easily simulated by lexicographic tie-breaking rules. To obtain them, it is sufficient to order the candidates with respec... |
4.1 Utilitarian and Candidate-Wise Egalitarian: Tie-Breaking Is Easy As a warm-up, we observe that tie-breaking can be applied and evaluated efficiently if the [MATH] -egroups are evaluated according to the utilitarian or candidate-wise egalitarian variant. The corresponding result follows almost directly from Proposit... |
Corollary 1 Let [MATH] denote the number of candidates and [MATH] denote the number of manipulators. Then one can solve [MATH] -Tie-Breaking in [MATH] time for [MATH] [MATH] |
Proof. The algorithm works in two steps. First, compute a lexicographic tie-breaking rule [MATH] that simulates [MATH] in [MATH] time as described in Proposition 1 . Second, apply tie-breaking rule [MATH] , and evaluate the resulting [MATH] -egroup in [MATH] time. The running time of applying a lexicographic tie-breaki... |
Subsection 3.3 ). 4.2 Egalitarian: Being Optimistic Is Hard In this subsection, we consider the optimistic and pessimistic tie-breaking rules when applied for searching a [MATH] -egroup evaluated according to the egalitarian variant. First, we show that applying and evaluating egalitarian tie-breaking is computationall... |
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