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Theorem 1 Let [MATH] denote the number of candidates, [MATH] denote the number of manipulators, [MATH] denote the evaluation threshold, and [MATH] denote the size of an egroup. Then one can solve [MATH] -Tie-Breaking in [MATH] time, but [MATH] -Tie-Breaking is [MATH] -hard and [MATH] -hard when parameterized by [MATH] ... |
Proof. For the pessimistic case, it is sufficient to “guess” the least satisfied manipulator [MATH] by iterating through [MATH] possibilities. Then, select |
[MATH] pending candidates with the smallest total utility for this manipulator in [MATH] time. Finally, comparing the [MATH] -egroup with the worst minimum satisfaction over all manipulators to the lower bound [MATH] on satisfaction level given in the input solves the problem. |
We prove the hardness for the optimistic case reducing from the [MATH] -hard Set Cover problem which, given a collection [MATH] of subsets of universe |
[MATH] and an integer [MATH] , asks whether there exists a family [MATH] of size at most [MATH] such that [MATH] . Let us fix an instance [MATH] of |
Set Cover . To construct an [MATH] -Tie-Breaking instance, we introduce pending candidates [MATH] representing subsets in [MATH] and manipulators [MATH] representing elements of the universe. Note that there are no confirmed and rejected candidates. Each manipulator [MATH] gives utility one to candidate [MATH] if set [... |
Observe that if there is a size- [MATH] subset [MATH] such that [MATH] , then there exists a family [MATH] —consisting of the sets represented by candidates in [MATH] —such that each element of the universe belongs to the set [MATH] . On the contrary, if we cannot pick a group of candidates of size [MATH] for which eve... |
[MATH] is a ‘no’ instance. This follows from the fact that for each size- [MATH] subset [MATH] there exists at least one manipulator [MATH] for whom [MATH] . This translates to the claim that there exists no size- [MATH] subset [MATH] such that all elements in [MATH] belong to the union of the sets in [MATH] |
Since Set Cover is [MATH] -hard and [MATH] -hard with respect to parameter [MATH] , we obtain that our problem is also [MATH] -hard and |
[MATH] -hard when parameterized by the size [MATH] of an excellence-group. Inspecting the [MATH] -hardness proof of Theorem 1 , we learn that a small egroup size (alone) does not make [MATH] -Tie-Breaking |
computationally tractable even for very simple utility functions. Next, using a parameterized reduction from the [MATH] -complete Multicolored Clique problem (Fellows et al., 2009 we show that there is still no hope for fixed-parameter tractability (under standard assumptions) even for the combined parameter “number of... |
Theorem 2 Let [MATH] denote the size of an egroup and [MATH] denote the number of manipulators. Then, parameterized by [MATH] [MATH] -Tie-Breaking is |
[MATH] -hard. Proof. We describe a parameterized reduction from the [MATH] -hard Multicolored Clique problem (Fellows et al., 2009 . In this problem, given an undirected graph [MATH] , a non-negative integer [MATH] , and a vertex coloring [MATH] , we ask whether graph [MATH] admits a colorful [MATH] -clique, that is, a... |
instance. Let [MATH] denote the set of vertices of color [MATH] , and let [MATH] , defined for [MATH] [MATH] , denote the set of edges that connect a vertex of color [MATH] to a vertex of color [MATH] |
Candidates. We create one confirmed candidate [MATH] and [MATH] pending candidates. More precisely: for each [MATH] , we create one vertex candidate |
[MATH] for each vertex [MATH] [MATH] and for each [MATH] such that [MATH] we create one edge candidate [MATH] for each edge [MATH] [MATH] We set the size [MATH] of the egroup to [MATH] and set the evaluation threshold [MATH] . Next, we describe the manipulators and explain the high-level idea of the construction. |
Manipulators and main idea. Our construction will ensure that there is a [MATH] -egroup [MATH] with [MATH] and [MATH] if and only if |
[MATH] contains [MATH] vertex candidates and [MATH] edge candidates that encode a colorful [MATH] -clique. To this end, we introduce the following manipulators. |
1. For each color [MATH] , there is a color manipulator [MATH] ensuring that the [MATH] -egroup contains a vertex candidate [MATH] |
corresponding to a vertex of color [MATH] . Herein, variable [MATH] denotes the id of the vertex candidate (resp. vertex) that is selected for color [MATH] |
2. For each [MATH] such that [MATH] , there is one color pair manipulator [MATH] ensuring that the [MATH] -egroup contains an edge candidate [MATH] corresponding to an edge connecting vertices of colors [MATH] and [MATH] . Herein, variable [MATH] denotes the id of the edge candidate (resp. edge) that is selected for co... |
[MATH] 3. For each [MATH] such that [MATH] , there are two verification manipulators [MATH] [MATH] ensuring that vertex [MATH] is incident to edge [MATH] if [MATH] or incident to edge [MATH] otherwise. |
If there exists a [MATH] -egroup in agreement with the description in the previous three points, then this [MATH] -egroup must encode a colorful [MATH] -clique. |
Utility functions. Let us now describe how we can guarantee correct roles of the manipulators introduced in points to above using utility functions. |
1. Color manipulator [MATH] [MATH] , has utility [MATH] for the confirmed candidate [MATH] , utility one for each candidate corresponding to a vertex of color [MATH] , and utility zero for the remaining candidates. |
2. Color pair manipulator [MATH] [MATH] [MATH] , has utility [MATH] for the confirmed candidate [MATH] , utility one for each candidate corresponding to an edge connecting a vertices of colors [MATH] and [MATH] , and utility zero for the remaining candidates. |
3. Verification manipulator [MATH] [MATH] [MATH] , has utility [MATH] for candidate [MATH] [MATH] , utility [MATH] for each candidate corresponding to an edge that connects vertex [MATH] |
to a vertex of color [MATH] , and utility zero for the remaining candidates. 4. Verification manipulator [MATH] [MATH] [MATH] , has utility [MATH] for candidate [MATH] [MATH] , utility [MATH] |
for each candidate corresponding to an edge that connects vertex [MATH] to a vertex of color [MATH] , and utility zero for the remaining candidates. |
Correctness. We argue that the graph [MATH] admits a colorful clique of size [MATH] if and only if there is a [MATH] -egroup [MATH] with [MATH] and [MATH] |
Suppose that there exists a colorful clique [MATH] of size [MATH] . Create the [MATH] -egroup [MATH] as follows. Start with [MATH] and add every vertex candidate that corresponds to some vertex of [MATH] and every edge candidate that corresponds to some edge of [MATH] . Each color manipulator and color pair manipulator... |
Suppose that there exists a [MATH] -egroup [MATH] such that [MATH] . Since each color manipulator cannot achieve utility [MATH] unless [MATH] |
belongs to the winning [MATH] -egroup, it follows that [MATH] . Because each color manipulator [MATH] receives total utility at least [MATH] [MATH] must contain some vertex candidate [MATH] corresponding to a vertex of color [MATH] for some [MATH] . We say that [MATH] |
selects vertex [MATH] . Since each color pair manipulator [MATH] receives total utility at least [MATH] [MATH] must contain some edge candidate [MATH] corresponding to an edge connecting a vertex of color [MATH] and a vertex of color [MATH] for some [MATH] . We say that |
[MATH] selects edge [MATH] . We implicitly assumed that each color manipulator and color pair manipulator contributes exactly one selected candidate to [MATH] . This assumption is true because there are exactly [MATH] such manipulators and each needs to select at least one candidate; hence, [MATH] is exactly of the des... |
Finally, devising an ILP formulation, we show that [MATH] -Tie-Breaking becomes fixed-parameter tractable when parameterized by the combined parameter “number of manipulators and number of different utility values.” This parameter covers situations with few manipulators that have simple utility functions; in particular... |
time. Theorem 3 Let [MATH] denote the number of different utility values and [MATH] denote the number of manipulators. Then, parameterized by [MATH] [MATH] -Tie-Breaking is fixed-parameter tractable. |
Proof. We define the type of any candidate [MATH] to be the size- [MATH] vector [MATH] . Let [MATH] be the set of all possible types. Naturally, the size of [MATH] is upper-bounded by |
[MATH] We denote the set of candidates of type [MATH] by [MATH] Now, the ILP formulation of the problem using exactly [MATH] variables reads as follows. For each type |
[MATH] , we introduce variable [MATH] indicating the number of candidates of type [MATH] in an optimal [MATH] -egroup. We use variable |
[MATH] to represent the minimal value of the total utility achieved by manipulators. We define the following ILP with the goal to maximize [MATH] |
(indicating the utility gained by the least satisfied manipulator) subject to: [EQUATION] Constraint set ( ) ensures that the solution is achievable with given candidates. Constraint ( ) guarantees a choice of an egroup of size [MATH] . The last set of constraints imposes that [MATH] holds at most the minimal value of ... |
is fixed-parameter tractable when parameterized by the combined parameter [MATH] Complexity of Coalitional Manipulation In the previous section, we have seen that breaking ties optimistically or pessimistically—an essential subtask to be solved by the manipulators in general—can be computationally challenging; in most ... |
[MATH] Input: An election [MATH] , a searched egroup size [MATH] [MATH] manipulators represented by their utility functions [MATH] |
such that [MATH] , and a non-negative, integral evaluation threshold [MATH] Question: Is there a size- [MATH] multiset [MATH] of manipulative votes over [MATH] such that [MATH] -egroup [MATH] wins the election [MATH] under [MATH] and |
[MATH] , and [MATH] [MATH] [MATH] [MATH] Coalitional Manipulation [MATH] [MATH] [MATH] CM The [MATH] [MATH] [MATH] CM problem is defined very generally; namely, one can consider any multiwinner voting rule [MATH] (in particular, any single-winner voting rule is a multiwinner voting rule with [MATH] ). In our paper, how... |
[MATH] [MATH] [MATH] CM to the [MATH] [MATH] [MATH] CM problem. In line with our intention to model optimistic and pessimistic attitudes of manipulators, we require that the evaluation of an optimistic/pessimistic tie-breaking rule [MATH] matches the manipulator’s evaluation. More formally, for every [MATH] , we focus ... |
On the way to show our results, we also use a restricted version of [MATH] [MATH] [MATH] Coalitional Manipulation that we call [MATH] [MATH] [MATH] Coalitional Manipulation |
with consistent manipulators . In this variant, the input stays the same, but all manipulators cast exactly the same vote to achieve the objective. |
To increase readability, we decided to represent manipulators by their utility functions. As a consequence, we frequently use, for example, [MATH] referring to the manipulator itself, even if we do not care about the values of utility function [MATH] at the moment of usage. In the paper, we also stick to the term “vote... |
As for the encoding of the input of [MATH] [MATH] [MATH] CM , we use a standard assumption; namely, that the number of candidates, the number of voters, and the number of manipulators are polynomially upper-bounded in the size of the input. Analogously to [MATH] -Tie-Breaking , both the evaluation threshold and the uti... |
5.1 Utilitarian & Candidate-Wise Egalitarian: Manipulation is Tractable We show that [MATH] [MATH] [MATH] Coalitional Manipulation can be solved in polynomial time for any constant [MATH] , any [MATH] , and any [MATH] . Whereas in general, for [MATH] being the input size, our algorithm requires [MATH] steps, for Bloc (... |
In several proofs in Subsection 5.1 we use the value of a candidate for manipulators (coalition) and say that a candidate is more valuable or less valuable than another candidate. Although we cannot directly measure the value of a candidate for the whole manipulators’ coalition in general, thanks to |
Observation 1 , we can assume a single non-zero utility function when discussing the utilitarian and candidate-wise egalitarian variants. Thus, assigning a single value to each candidate is justified. |
We start with an algorithm solving the general case of [MATH] Bloc [MATH] [MATH] Coalitional Manipulation [MATH] [MATH] . The basic idea is to “guess” the lowest final score of a member of a [MATH] -egroup and (assuming some lexicographic order over the candidates) the least preferred candidate of the [MATH] -egroup th... |
Theorem 4 Let [MATH] denote the number of candidates, [MATH] the number of voters, [MATH] the size of a searched egroup, and [MATH] the number of manipulators. One can solve [MATH] Bloc [MATH] [MATH] Coalitional Manipulation in |
[MATH] time for any [MATH] and [MATH] Proof. We prove the theorem for the lexicographic tie-breaking rule [MATH] . This is sufficient since, using Proposition 1 , one can generalize the proof for the cases of utilitarian and candidate-wise egalitarian variants. The basic idea of our algorithm is to fix certain paramete... |
problem with polynomial-sized weights. The algorithm iterates through all possible value combinations of the following two parameters: |
the lowest final score [MATH] of any member of the [MATH] -egroup and the candidate [MATH] that is the least preferred member of the |
[MATH] -egroup with final score [MATH] with respect to tie-breaking rule [MATH] Having fixed [MATH] and [MATH] , let [MATH] denote the set of candidates who get at least [MATH] approvals from the non-manipulative voters or who are preferred to [MATH] according to [MATH] and get exactly [MATH] approvals from the non-man... |
[MATH] -egroup. Let [MATH] . For sanity, we check whether [MATH] that is, whether candidate [MATH] can belong to the [MATH] -egroup if the candidate obtains final score [MATH] . We discard the corresponding combination of solution parameter values if the check fails. Next, we ensure that [MATH] |
obtains the final score exactly [MATH] . If [MATH] receives less than [MATH] or more than [MATH] approvals from non-manipulative voters, then we discard this combination of solution parameter values. Otherwise, let [MATH] denote number of additional approvals candidate [MATH] needs in order to get final score [MATH] . ... |
[MATH] be the number of remaining (not yet fixed) members of the [MATH] -egroup. Let [MATH] be the number of approvals to be distributed to candidates in [MATH] |
Now, the manipulators have to influence further [MATH] candidates to join the [MATH] -egroup (so far only consisting of [MATH] ) and distribute exactly [MATH] approvals in total to candidates in [MATH] but at most [MATH] approvals per candidate. To this end, let [MATH] denote the set of candidates which can possibly jo... |
1. [MATH] if [MATH] is preferred to [MATH] with respect to [MATH] , or 2. [MATH] if [MATH] is preferred to [MATH] with respect to [MATH] |
A straightforward idea is to select the [MATH] elements from [MATH] which have the highest values (that is, utility) for the coalition. However, there can be two issues: First, [MATH] might be too small; that is, there are too few approvals to ensure that each of the [MATH] best-valued candidates gets the final score a... |
Fortunately, we can easily detect these cases and deal with them efficiently. In the former scenario we reduce the remaining problem to an instance of |
Exact [MATH] -item Knapsack —the problem in which, for a given set of items, their values and weights, and a knapsack capacity, we search for [MATH] |
items that maximize the overall value and do not exceed the knapsack capacity. In the latter case, we show that we can discard the corresponding combination of solution parameters. |
First, if [MATH] , then one can certainly distribute all [MATH] approvals (e.g., to the [MATH] candidates that will finally join the |
[MATH] -egroup). Of course, it could still be the case that there are too few approvals available to push the desired candidates into the [MATH] -egroup in a greedy manner. To solve this problem, we build an Exact [MATH] -item Knapsack instance where each candidate in [MATH] is mapped to an item. We set the weight of e... |
[MATH] with respect to [MATH] and otherwise to [MATH] . We set the value of each [MATH] to be equal to the utility that candidate [MATH] |
contributes to the manipulators. Now, an optimal solution (given the combinations of parameter values is correct) must select exactly [MATH] elements from [MATH] such that the total weight is at most [MATH] . This corresponds to Exact |
[MATH] -item Knapsack if we set our knapsack capacity to [MATH] . Furthermore, finding any such set with maximum total value leads to an optimal solution. Even if the final total weight [MATH] of the chosen elements is smaller than |
[MATH] , we can transfer the Exact [MATH] -item Knapsack solution to the correct solution of our problem. The total weight corresponds to the number of approvals used. Thus, with the Exact [MATH] -item Knapsack solution we spend [MATH] approvals and, because of the monotonicity of [MATH] Bloc together with the assumpti... |
approvals to approve the chosen candidates even more. Second, if [MATH] , then one can certainly ensure for any set of [MATH] candidates from [MATH] the final score at least [MATH] (resp. at least |
[MATH] ). In many cases, it will not be a problem to distribute the approvals; for example, one can safely spend up to [MATH] approvals for each candidate from [MATH] , that is, to candidates that have no chance to get enough points to join the [MATH] -egroup or to candidates which are already fixed to be in the [MATH]... |
into the [MATH] -egroup (spending [MATH] approvals) and then safely distribute the remaining approvals within [MATH] as discussed. If [MATH] , then there is no possibility of distributing approvals in a way that [MATH] is part of the [MATH] -egroup. Towards a contradiction let us assume that [MATH] is part of the |
[MATH] -egroup obtained after distributing [MATH] approvals. This means that we spend all possible [MATH] approvals so that [MATH] is not beaten and [MATH] approvals to push [MATH] candidates to the winning [MATH] -egroup. Giving one more approval to some candidate [MATH] from [MATH] that is not yet in the [MATH] -egro... |
[MATH] out of the final [MATH] -egroup; a contradiction. Consequently, for the case of [MATH] , we discard the corresponding combination of solution parameters. |
As for the running time, the first step is sorting the candidates according to their values in [MATH] time. Then let us consider the running time of two cases [MATH] and [MATH] separately. In the former case, we solve Exact [MATH] -item Knapsack in [MATH] time by using dynamic programming based on analyzing all possibl... |
[MATH] , then we approve at most [MATH] candidates which gives the running time [MATH] . Thus, we can conclude that the running time of the discussed cases is [MATH] . Additionally, there are at most |
[MATH] values of [MATH] and at most [MATH] values of [MATH] . Summarizing, we get the running time [MATH] Next, we show that, actually, Bloc [MATH] [MATH] CM (i.e., a special case of [MATH] [MATH] [MATH] CM where [MATH] ) can be solved in quadratic-time, that is, in practice, much faster than the general variant of the... |
Proposition 3 Let [MATH] denote the number of candidates, [MATH] denote the number of voters, and [MATH] denote the number of manipulators. Then one can solve |
[MATH] [MATH] [MATH] Coalitional Manipulation with consistent manipulators in [MATH] time for any [MATH] and [MATH] Proof. Consider an instance of [MATH] [MATH] [MATH] CM with consistent manipulators with an election [MATH] |
where [MATH] is a candidate set and [MATH] is a multiset of non-manipulative votes, [MATH] manipulators, an egroup size [MATH] , and a lexicographic order |
[MATH] used by [MATH] to break ties. In essence, we introduce a constrained solution form called a canonical solution and argue that it is sufficient to analyze only this type of solutions. Then we provide an algorithm that efficiently seeks for an optimal canonical solution. |
At the beginning, we observe that when manipulators vote consistently, then we can arrange the top [MATH] candidates of a manipulative vote in any order. Hence, the solution to our problem is a size- [MATH] subset (instead of an order) of candidates which we call a set of supported candidates ; we call each member of t... |
Strength order of the candidates. Additionally, we introduce a new order [MATH] of the candidates. It sorts them descendingly with respect to the score they receive from voters and, as a second criterion, according to the position in [MATH] Intuitively, the easier it is for some candidate to be a part of a winning |
[MATH] -egroup, the higher is the candidate’s position in [MATH] . As a consequence, we state Claim 1 Claim 1 Let us fix an instance of [MATH] [MATH] [MATH] CM with consistent manipulators and a solution [MATH] which leads to a winning [MATH] -egroup [MATH] . For every supported (resp. unsupported) candidate [MATH] , t... |
1. If [MATH] is part of the winning [MATH] -egroup, then every supported (resp. unsupported) predecessor of [MATH] , according to [MATH] , belongs to [MATH] |
2. If [MATH] is not part of the winning [MATH] -egroup, then every supported (resp. unsupported) successor of [MATH] , according to [MATH] , does not belong to [MATH] |
Proof. Fix an instance of [MATH] [MATH] [MATH] CM with consistent manipulators, a solution [MATH] , and a winning [MATH] -egroup [MATH] . Let us consider the respective order [MATH] over the candidates in the instance. |
We first show that statement regarding supported candidates holds. According to the statement, fix some supported candidate [MATH] and let [MATH] be a predecessor of [MATH] (according to [MATH] ). Towards a contradiction, let us assume that [MATH] . This implies that either (i) the score of [MATH] is smaller than the s... |
An analogous approach leads to proofs for the remaining three cases stated in the theorem.∎ Claim 1 justifies thinking about [MATH] as a “strength order”; hence, in the proof we use the terms stronger and weaker |
candidate. Using Claim 1 , we can fix some candidate [MATH] as the weakest in the winning [MATH] -egroup and then infer candidates that have to be and that cannot be part of this [MATH] -egroup. To formalize this idea, we introduce the concept of a canonical solution |
Canonical solutions. Assuming the case where [MATH] , we call a solution [MATH] leading to a winning [MATH] -egroup [MATH] canonical if all candidates of the winning egroup are supported; that is, [MATH] . In the opposite case, [MATH] solution [MATH] is canonical if [MATH] and [MATH] is a set of the [MATH] |
weakest candidates in [MATH] . For the latter case, the formulation describes the solution which favors supporting weaker candidates first and ensures that no approval is given to a candidate outside the winning [MATH] -egroup. |
Canonical solutions are achievable from every solution without changing the outcome. Observe that one cannot prevent a candidate from winning by supporting the candidate more because this only increases the candidate’s score. Consequently, we can always transfer approvals to all candidates from the winning [MATH] -egro... |
[MATH] -egroup. Dropped and kept candidates. Observe that for every solution (including canonical solutions), we can always find the strongest candidate who is not part of the winning egroup. We call this candidate the dropped candidate . Note that we use the strength order in the definition of the dropped candidate; t... |
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