On the Hardness of Bribery Variants in Voting with CP-Nets
aa r X i v : . [ c s . G T ] M a y On the Hardness of Bribery Variants in Votingwith CP-Nets ∗ Britta Dorn and Dominikus Krüger Faculty of Science, Dept. of Computer Science, University of Tübingen, Germany, [email protected] Dominikus Krüger, Institute of Theoretical Computer Science, Ulm University,Germany, [email protected]
Abstract
We continue previous work by Mattei et al. [41] in which they studythe computational complexity of bribery schemes when voters have con-ditional preferences modeled as CP-nets. For most of the cases they con-sidered, they showed the bribery problem is solvable in polynomial time.Some cases remained open—we solve several of them and extend the previ-ous results to the case that voters are weighted. Additionally, we considernegative (weighted) bribery in CP-nets, when the briber is not allowed topay voters to vote for his preferred candidate.
This work is based on the findings of Mattei, Pini, Rossi, and Venable [41].They considered the computational complexity of computing optimal briberyschemes in voting scenarios in which the voters decide on a fixed assignmentfor a common set of issues for which they might have conditional preferences.The typical example for this setting is the choice of a common meal consistingof several courses and drinks. Here it is a natural assumption that some voters’preference over their choice of wine is conditioned on the kind of meat beingserved, which again might depend on the choice of previously chosen dishes.In such a setting, the set of candidates to vote on is exponentially large in thenumber of common issues. A compact and convenient way to condense andrepresent the voters’ possibly conditional preferences over this set is given bythe CP-net formalism introduced by Boutilier et al. [5], see also earlier worksby Boutilier et al. [4, 6]. ∗ The final publication is available at Springer via http://dx.doi.org/10.1007/s10472-015-9469-3 .
1n a CP-net, a voter’s dependencies for a set of issues are modelled by adirected graph, and the conditional aspect of his preferences is expressed byso-called ceteris paribus conditional preference statements for each issue.Preference aggregation in CP-nets was first investigated by Rossi et al. [47]and was further pursued in various works such as Purrington and Durfee [46],Li et al. [36, 37], Xia et al. [50], Conitzer et al. [9], to name just a few. Amongthe approaches for voting with CP-nets, there are one-step methods that, giventhe voters’ CP-nets, derive a linear preference order (or relevant parts of it) overthe candidates and apply one of the existent rules for elections where prefer-ences are given as linear orders. Lang [33] proposed a sequential approach whichaggregates preferences issue by issue, see also the work of Lang and Xia [34] andXia et al. [51, 52]. In their work, Mattei et al. [41] use the one-step meth-ods One-Step-Plurality ( OP ), One-Step-Veto ( OV ), and One-Step- k -Approval( OK ) (and a variant called OK ∗ ) which are derived from the well-known vot-ing rules for linear preference orders, as well as the sequential voting systemSequential Majority ( SM ).A central challenge in the field of computational social choice consists in de-termining the computational complexity of voting problems [7, 25]. One seeksto assess the extent to which voting systems are—in some sense—vulnerable toor resistant against manipulative actions such as strategic behavior (manipula-tion), election control, or bribery. These properties are measured in terms ofcomputational hardness of the corresponding decision problems. For the defini-tion of these problems and for surveys on this topic we refer to the article byFaliszewski et al. [25] and the bookchapter by Faliszewski et al. [22] on manip-ulation, bribery, and control, the article by Faliszewski and Procaccia [26] onmanipulation, the survey by Rothe and Schend [48] on manipulation and con-trol, and to the bookchapters by Conitzer and Walsh [10] for the manipulationproblem, and by Falizewski and Rothe [27] for bribery and control.In this work, we consider the bribery problem which asks whether an externalagent, the briber, can influence the voters by spending money on changingtheir preferences over the candidates in such way that his preferred candidatewins, respecting a given budget. This problem was introduced and intensivelystudied by Faliszewski et al. [21]. The original version deals with individualbut fixed prices for each voter, independent of the specific changes made ifthe voter is bribed. Several variants have been considered [21], among themthe model of nonuniform bribery introduced by Faliszewski [19] and the modelof microbribery studied by Faliszewski et al. [23] which may incorporate theamount of change the briber asks for. The Swap Bribery problem introducedby Elkind et al. [17] additionally takes into account the ranking aspect of thevoters’ preferences by assigning costs for swapping two consecutive candidatesin a preference order.Given that bribery of an election represents some kind of dishonest behavior,one is interested in deriving hardness results for voting systems with respect tothe computational complexity of the bribery problem, even if this only consti-tutes a worst case analysis. 2owever, the bribery problem can also be considered in a positive way interms of convincing voters by (possibly) cost-involving actions like campaigningto change their votes, and is referred to as election campaign management then,see the works of Elkind et al. [17], Elkind and Faliszewski [16], Schlotter etal. [49], and Baumeister et al. [2].Considering the bribery problem as a campaign management problem, one isof course less interested in obtaining computational hardness than in findingefficient polynomial time algorithms.In this work, we further investigate the complexity of the bribery problemin the setting of voting with CP-nets as initiated by Mattei et al. [41]. Thedifference to the ‘classical’ setting where preferences are given as linear ordersis that the briber does not directly execute changes in the preference orders ofthe voters, but in the CP-nets, i.e., he can affect the dependencies. From thepoint of view of campaign management, this seems very natural as well. Onecan easily imagine a systematic campaign to convince voters to drop dependen-cies in their preferences, without having to deal with their implicit preferenceorder. For instance, if a voter prefers some special kind of wine in the case thatmeat is served as a main dish, one might run a campaign for the quality of thealternative varieties of wine offered and therewith directly affect the CP-net,without having to care about the induced preference order.Bribery in CP-nets has also been investigated by Maran et al. [38] in the contextof interaction and influence among voters and by Pini et al. [45] in connectionto representation of the voters’ preferences via soft constraints.In the ‘classical’ setting of unconditional preferences, the basic bribery prob-lem is solvable in polynomial time for the voting rule k -approval [21], and there-fore also for its special cases plurality and veto. In contrast, the Swap Bribery problem for k -approval is solvable in polynomial time if all swaps have a costof one, but becomes N P -complete as soon as different costs are allowed, al-ready for the case k = 2 (see the work of Dorn and Schlotter [13], Betzler andDorn [3] and Elkind et al. [17]); the complexity hence depends—additionally tothe voting system used—on the amount of change that the briber has to payfor.For bribery in the CP-net setting, the amount of change that the briberhas to pay for is incorporated by a cost scheme. Mattei et al. introduce fivecost schemes, called C equal , C flip , C level , C dist , and C any , which are inspiredby the classical setting, ranging from unitary costs irrespective the amount ofchange per voter, over the scenario of swap bribery where specific costs forswaps in the preferences can be modelled, to the case where arbitrary costcan be incorporated. Additionally, these cost schemes are extended by a costvector Q ∈ ( N ) n which allows for modelling an individual cost factor for eachvoter. Also, different bribery actions are considered. Mattei et al. distinguishthe cases that the briber can affect changes in so-called independent variablesonly (IV), meaning that he can only make changes concerning issues for whichthe preferences are independent of the outcome for other issues, in dependentvariables only (DV), or in all kinds of variables (IV+DV).3n most of the cases they considered, Mattei et al. obtained that the briberyproblem is easy, i.e., solvable in polynomial time. An overview of their resultsis given in Table 1. Three cases with different variants remained unanswered:The complexity for the voting systems OP , OV , OK ∗ with the cost scheme C level for bribery actions DV and IV + DV , the complexity for the votingsystems OP , OV , OK ∗ with the cost scheme C any for all bribery actions, andthe complexity for SM with the cost scheme C dist for all bribery actions. Thefirst two cases were believed to be N P -complete. We show that these two aresolvable in polynomial time as well.Moreover, we extend Mattei et al. ’s results for the case that voters areweighted. So far, the weighted version of the bribery problem in CP-netswas considered for the voting system SM only. Mattei et al. obtained N P -completeness for all cost schemes and all bribery actions in this case. Addi-tionally, for the cost scheme C flip , they showed that the problem is solvablein polynomial time if no individual voter costs are allowed, see Table 1. Wecomplement their results by proving N P -completeness for most variants of theweighted version for the other voting rules as well, and further illuminate theinfluence of individual voter costs for the complexity of the weighted versions.We find that considering individual extra costs for voters only makes a differ-ence in terms of complexity for the voting system weighted SM combined withthe cost schemes C flip and C level .Additionally, we investigate the case of negative bribery, a variant introducedby Faliszewski et al. [20], where it is not allowed to pay voters to vote for thepreferred candidate, both for the unweighted and weighted versions. We obtainthat the unweighted negative bribery problem is likewise solvable in polynomialtime for almost all variants considered so far, except for SM combined with C equal where it is N P -complete, and for SM combined with C dist , which stillis unsolved, like its correspondent in the positive version. The weighted nega-tive version is N P -complete for all variants considered so far. Our results aresummarized in Table 2 in Section 4.This work is organized as follows. In Section 2, we give the basic notions usedfrom CP-nets, voting, and bribery. We introduce the setting of Mattei et al. ,the definitions of the bribery variants to be investigated, and the
N P -completeproblems that we use in our reductions. Our results are contained in Section 3.We start with proving Theorem 2 for finding the ‘cheapest subsets’ of a finiteset which helps in solving some of the open problems from Mattei et al. [41]in the unweighted variants of the bribery problem. The subsequent subsectionsdeal with the complexity of the weighted case, where reductions from the
N P -complete
Partition problem are given, and investigations for the unweightedand weighted cases of negative bribery. An overview of our results and openproblems is provided in Section 4, together with a discussion and directions forfuture research. 4able 1: Complexity results obtained by Mattei et al. [41] for the bribery prob-lem in CP-nets. P stands for solvability in polynomial time, N P -c for
N P -completeness. The given results hold for the bribery actions IV , DV , and IV + DV . Questions that were unsolved by Mattei et al. are marked with‘ ?’. The case of weighted SM with C flip labeled with two complexity classesmeans that the problem can be solved in polynomial time if no individual votercosts are allowed, and that it is N P -complete if individual costs are taken intoaccount. C equal C flip C level C any C dist SM N P -c P P P ?OP
P P P for IV / ? for DV , IV + DV ? P OV P P P for IV / ? for DV , IV + DV ? P OK*
P P P for IV / ? for DV , IV + DV ? P weighted SM N P -c P , N P -c N P -c N P -c N P -c We mostly use notation and definitions as introduced by Mattei et al. [41].
In our setting, we are given a set of m issues M = { X , . . . , X m } each of whichhas a binary domain D ( X j ) = { x j , x j } , where j ∈ { , . . . , m } . A completeassignment to all issues is called an outcome or a candidate , the set of candidates D ( X ) × · · · × D ( X m ) hence consists of m elements. Each of the n voters has(possibly) conditional preferences over the values assigned to the issues: thevalue assigned to an issue by a voter might affect the values of other issues.In this case, the latter issues are called dependent issues; an issue is called independent if its value does not depend on the value of another issue. Thedependencies can be modeled by a directed dependency graph having vertexset M and a directed edge going from X i to X j if and only if the assignmentof the value of X j depends on the assignment of the value of X i . A CP-net over M consists of a dependency graph together with a conditional preferencetable for each issue X i (or vertex of the dependency graph, respectively) wherethe voter specifies a strict total order over the values of X i for each completeassignment of the issues on which it depends on. Each of these specifications isreferred to as a cp-statement . For example, if for an independent issue A , theassignment A = a is unconditionally preferred to A = a , the cp-statement iswritten as a > a ; if the assignment A = a is preferred to A = a , we write a > a .For a set of issues { A, B } with domains D ( A ) = { a, a } and D ( B ) = { b, b } inwhich B is dependent on A , if a is unconditionally preferred over a , and b ispreferred over b in case the value of a is assigned to A , and b is preferred over b in case the value of a is assigned to A , we have one cp-statement for A , namely5 > a , and two cp-statements for B , which we write as a : b > b,a : b > b. A CP-net is called acyclic if the corresponding dependency graph is acyclic;it is called compact if the number of issues each issue depends on is boundedby a constant. We remark that the original definition of CP-nets by Boutilier et al. [5] does not require acyclicity, but this assumption is natural [33] andalso employed by Mattei et al. [41]. In acyclic CP-nets, there is only one mostpreferred candidate which can then be found in linear time by going throughthe CP-net in topological order and assigning the most preferred value to eachissue according to the cp-statements [5].Throughout this work, we assume that the voters’ preferences on the set ofissues are given by acyclic and compact CP-nets. The collection of CP-nets ofall voters is called a profile . Example 1.
A group of friends wants to go on vacation. They can choosebetween going to Austria or Italy for hiking or alpine skiing in the summer orthe winter holidays. The three issues to decide on are hence
Where , When ,and
What . So the corresponding domains are D ( Where ) = { Austria , Italy } , D ( When ) = { summer , winter } , and D ( What ) = { skiing , hiking } . A possiblecandidate would be skiing in summer in Italy . Figure 1 shows two possibleconditional preferences modelled as CP-nets, each of them consisting of a de-pendency graph and a conditional preference table.
CP-nets only define a partial order over the candidates, i.e., some candidatesare incomparable. There are several ways to create strict total orders over thecandidates ([7, 11]). We use the linearization method that is also used by Mat-tei et al. : Each voter provides a fixed strict total order X i > X i > · · · > X i m over the issues X , . . . , X m such that each issue is independent from all issuesfollowing it in this order; this is possible because the CP-net is acyclic. Then weassociate a binary vector of length m to each candidate, for which the entry atposition j corresponds to issue X i j in the voter’s fixed ordering. This entry is setto if the preferred value of its binary domain is assigned to it, and otherwise.Hence, the most preferred candidate is associated to the vector (0 , . . . , , andthe least preferred one to the vector (1 , . . . , . Given a candidate, the next bestcandidate can be found efficiently by increasing the binary number representedby its vector by one.In the general setting, every voter may have an individual fixed total orderover the issues. The voting system SM (defined below) relies on a sequentialapproach to voting and therefore requires existence of a strict total order O overthe issues such that for each CP-net, each issue is independent from all issuesfollowing it in O . A profile that fulfills this property is called an O - legal profileby Lang [33], see also the work of Lang and Xia [34]. Mattei et al. [41] use thisnotion as well as the notion of a constant linearization scheme across all agents Where , When , and
What . The possiblevalues are Austria or Italy in the summer or the winter holidays for hiking oralpine skiing. Alice thinks that a summer vacation always is the better choice,but if it has to be winter she prefers alpine skiing in Austria. Bob thinks thatalpine skiing is not an appropriate summer activity and that only Italy has goodski regions.for such an order O . In Example 1, the profile consisting of Alice’s and Bob’sCP-nets is O -legal for the order O : When > Where > What . Given a profile of CP-nets over a set of issues, one can determine the winningoutcome(s) by a voting rule which maps a profile to a set of candidates. In the unique winner model, the rule determines a single winning candidate, whereasin the non-unique or co-winner model, the output of the rule is a whole set ofcandidates which are all considered as winners. In this case, the notion of avoting correspondence is also used for the notion of a voting rule. We considerthe same voting rules for CP-nets as Mattei et al. [41]. • Sequential majority ( SM ): Given a total order O for which the profile is O -legal, we follow this order issue by issue, and execute a majority votefor each issue. The voters fix the winning value of the corresponding issue7n their CP-net and then go on to the next issue. The (unique) winningcandidate is the combination of the winners of the individual steps taken. • One-Step- k -Approval ( OK ): The k most preferred candidates of eachvoter obtain point each, the remaining candidates obtain points. The(co-)winners of the election are the candidates with the maximum numberof points. In particular, we consider the two following special cases: – One-Step-Plurality ( OP ), where k = 1 , i.e., only the most preferredcandidate of each voter has to be considered, and – One-Step-Veto ( OV ), where k = 2 m − , i.e., only the least preferredcandidate of each voter has to be provided.For OK , O -legality of the profile is not necessarily required. Mattei et al. alsoconsider the special case of OK , denoted by OK ∗ , where k is a power of two.For more details and several examples for the use of these rules, we refer toMattei et al. [41]. We consider the problem that an external agent, the briber , who knows the CP-nets of all voters, asks them to execute changes in their cp-statements. Mattei etal. [41] define this in a way that the briber can ask the voters to flip the value ofone or more issues in their CP-nets, which might imply several further changes,according to the dependencies. They distinguish the case that the briber canask for a change in the cp-statements of the independent issues only (IV), thedependent issues only (DV), or in any cp-statement (IV+DV).Moreover, they introduce five cost schemes: • C equal , where any amount of change in the CP-net has the same unit cost, • C flip , where the cost of changing a CP-net is the total number of individualcp-statements that must be flipped to obtain the desired change, • C level , where the cost to flip a cp-statement is linked to the depth of theassociated issue in the dependency graph.The cost of a bribery is computed as P X j ∈ M flip ( X j ) · ( k + 1 − level ( X j )) ,where k is the number of levels in the CP-net, level ( X j ) corresponds tothe depth of issue X j in the dependency graph, and flip ( X j ) is the numberof flips performed in cp-statements associated with X j . (Note that for thevoting rules we consider, we have flip ( X j ) ∈ { , } for all j , see the remarkbelow.) More precisely,level ( X j ) = if X j is an independent issue; i + 1 if level ( X i ) ≤ i holds for all parents X i of X j andthere is at least one parent X l with level ( X l ) = i. C any , where the cost is the sum of the flips, each weighted by a specificcost, and • C dist , where we require a fixed order of the issues for each voter (but notnecessarily the same for each of them), which induces a strict total orderover all candidates. The cost to bribe a voter to make candidate c his topcandidate is the number of candidates who are better ranked than c inthis order. Remark.
We remark that for the considered voting rules and for each reason-able bribery, flip ( X j ) in the definition of C level is equal to or to for eachissue X j . We call a bribery reasonable if there is no lower priced bribery thathas the same benefit for the briber. Clearly, flip ( X ) ∈ { , } holds for an in-dependent issue X because there is only one cp-statement associated with anindependent issue. All dependent issues have more than one cp-statement, butfor the four voting rules we consider, it is never reasonable to bribe for changesin more than just one cp-statement per issue, as those additional changes maygenerate additional costs but are of no help. • For SM , when the majority vote is executed for an issue X , this meansthat for each voter all the issues that X depends on already have a fixedvalue. Therefore only one cp-statement per issue is relevant. • For OP , the briber only has to change the most preferred candidate of avoter in case he decides to bribe him. This is achieved by changing thevalue of some of the issues in the corresponding CP-net, and this can bedone by flipping a single cp-statement in the issues that are concerned. • For OV , this is almost the same. The briber only has to change the leastpreferred candidate, which means that the value of some issues has to beflipped, which can be done by flipping a single cp-statement per issue. • For OK , we have a set of approved candidates and a set of disapprovedcandidates. There is at most one cp-statement per issue which implies achange of these two sets if it is flipped. A flip of the other cp-statementsonly changes the order of the candidates within these sets. Additionally, these cost schemes are extended by a cost vector Q ∈ ( N ) n whichallows for modelling an individual cost factor for each voter. The factor forvoter v i is denoted by Q [ i ] and is multiplied with the costs calculated by a cer-tain cost scheme to obtain the amount that the briber has to pay to v i . Weremark that all our tractability results, except the one of Theorem 12, hold forarbitrary cost vectors, while all of our hardness results still hold with Q [ i ] = 1 for all ≤ i ≤ n .The ( D, A, C ) -bribery problem is then defined in the following way:9 D, A, C ) -Bribery Given:
A profile of n CP-nets over m common binary issues, awinner determination voting rule D ∈ { SM, OP, OV, OK } , a briberyaction A ∈ { IV, DV, IV + DV } , a budget β , a cost scheme C ∈{ C equal , C flip , C level , C any , C dist } , a cost vector Q ∈ ( N ) n , and apreferred candidate p . For the voting rule SM, we also require O -legality for a given total order O over the issues. Question:
Is it possible to bribe the voters to make changes intheir cp-statements in such a way that p becomes a (co-)winner ofthe bribed election, without exceeding β ?In this work, we focus on the non-unique winner model to keep the proofs assimple as possible. All our results still hold for the unique winner model. Ourproofs can easily be adapted for this model, e.g. by adding a tie-breaking voterwho initially votes for p , cf. Faliszewski et al. [21], or some similar adjustment.Note that in general, tie-breaking is not an easy task and an interesting questionof its own ([24, 40, 43, 44]).We also consider Weighted- ( D, A, C ) -Bribery , which is defined in thesame way, but with weighted voters, which is a typical variant for bribery prob-lems (see the overview of Faliszewski et al. [21]). Moreover, we investigate thecomputational complexity of weighted and unweighted ( D, A, C ) -Negative-Bribery , which are defined as their non-negative versions, but where the briberis not allowed to bribe voters to vote for his preferred candidate, see also Fal-iszewski et al. [21], who introduced this variant in the original setting of thebribery problem.For the polynomial time algorithm we give in Theorem 12, we consider thebribery problem as the optimization version of the Knapsack problem [30],which is defined as follows.
Knapsack
Given:
A finite set U , for each u ∈ U a size s ( u ) ∈ Z + and a value v ( u ) ∈ Z + , and a positive integer S . Task:
Find a subset U ′ ⊆ U such that P u ∈ U ′ s ( u ) ≤ S and P u ∈ U ′ v ( u ) is maximal.The Knapsack problem is known to be
N P -complete and can be solved intime O ( n · S ) or O ( n · T ) , where T := P u ∈ U v ( u ) , with a dynamic programmingalgorithm, see the work of Dantzig [12]. This running time is pseudo-polynomialsince it depends on the representation of the integers S or T . We will see that inour case, however, T is bounded by n · m . Neither n nor m can be exponentialin the input size, since each of the n voters is given by his CP-net and eachsuch CP-net consists of at least m cp-statements. This will lead to solvabilityin polynomial time with respect to the input size.To show N P -completeness for variants of the bribery problem, we use reductionsfrom the following problems. 10 artition
Given:
A set A of (not necessarily different) integers with P a ∈A a = 2 ψ . Question:
Is there a subset A ′ ⊂ A such that P a ∈A ′ a = ψ ?This problem is one of Karp’s first 21 N P -complete problems [31].The following problem is used for the reductions in the negative case.
Negative Optimal Lobbying
Given: An n × m / matrix E , a positive integer k , and a / vector x of length m . (Each row of E represents an agent. Eachcolumn represents a referendum in the election or a certain issue tobe voted on by the legislative body. The / values in a given rowrepresent the natural inclination of the agent with respect to thereferendum questions put to a vote in the election. The vector x represents the outcomes preferred by The Lobby.) Parameter: k (representing the number of agents to be influenced) Question:
Is there a choice of k rows of the matrix, such that theserows can be edited, without setting them equal to x , so that in eachcolumn of the resulting matrix, a majority vote in that column yieldsthe outcome targeted by The Lobby ( = x )?This problem is the negative version of the N P -complete
Optimal Lobbying problem [8]. As the following theorem states, it is
N P -complete as well. In fact,
Optimal Lobbying is even W [2] -hard with respect to the number k of agentsto be influenced. This number (which is part of the input) is called a parameter in the context of parameterized complexity theory. The class W [2] belongs to theso-called W -hierarchy within this theory. We refer to Downey and Fellows [14]for these definitions, or the books of Flum and Grohe [28] or Niedermeier [42]. Toprove membership and hardness with respect to the classes of the W -hierarchy,so-called parameterized reductions are used which have to meet certain require-ments on the parameters that are considered. The polynomial-time reductionwe give in the proof of the following theorem preserves the parameter k andtherefore is a simple case of a parameterized reduction, implying W [2] -hardnesswith respect to k for Negative Optimal Lobbying . Informally speaking, thismeans that the problem is not expected to be solved efficiently even for smallvalues of k . Theorem 1.
Negative Optimal Lobbying is N P -complete and W [2] -hardwith respect to the number of agents to be influenced.Proof. We give a reduction from
Optimal Lobbying . Starting with an in-stance I OL of Optimal Lobbying with an n × m / matrix E , a positiveinteger k , and a / vector x of length m , we construct the instance I NOL of Negative Optimal Lobbying by extending the vector x of the preferred out-come of The Lobby by n positions, all of them set to . The matrix E is extendedby n additional columns as well. We use the n × n identity matrix for this ex-tension. Therefore exactly one entry is in each of the additional columns, and11ach row has one -entry in exactly one new column. All remaining entries inthis extension are set to . The parameter k is left unchanged.We make two observations: First, the new matrix still has n rows, so themajority in the new columns is always . And second, there is exactly one ineach row in the new columns. So there is no need for the Lobby to edit anythingin the new columns and no row has to be changed to x . It is not hard to verifythat I OL is a yes-instance if and only if I NOL is a yes-instance.
In this section, we address several cases left open by Mattei et al. [41]. Theyshowed that the unweighted bribery problem is solvable in polynomial time forthe one-step voting systems OP , OV , OK ∗ for several cost schemes. For thecost scheme C any and all bribery actions, and for the cost scheme C level withbribery actions DV and IV + DV , the computational complexity was still open.In this section, we show that all of these problems are polynomially tractable aswell, thus completing the complexity table for OP , OV , OK ∗ in the unweightedcase. An overview of all results (both from Mattei et al. [41] and ours) can befound in Table 2 at the end of this work.One task in the proofs of Theorems 3 and 7 which address the open problemsconsists in finding the n cheapest candidates in the exponentially large set ofcandidates. We give a general formulation of this problem in terms of findingthe “smallest subsets” of a set as follows. By P ( A ) , we denote the power set ofa finite set A . K-Smallest Subsets
Given:
A finite set A = { a , a , . . . , a m } , size s ( a i ) ∈ Z + for ≤ i ≤ m , and a unary coded integer K ∈ N , K ≤ m = |P ( A ) | . Wanted:
The “ K smallest subsets” of A , i.e., the first K elementsof the power set P ( A ) , when its elements are sorted ascendingly (tiesbroken arbitrarily) by the sums of the sizes of their elements.In the following, by the size of a subset, we refer to the sum of the sizes ofits elements. The empty set has size zero. The K-Smallest Subsets henceasks for the K elements of P ( A ) with the smallest size. Example 2.
Let A = { a , . . . , a } with s ( a ) = s ( a ) = 1 , s ( a ) = 2 , s ( a ) =3 , s ( a ) = 4 , s ( a ) = 7 . For a better readability, we use the notation A = { , ˆ1 , , , , } in this example, where we mark one with a hat to be able todistinguish between a and a . Then the seven smallest subsets of A are the sets { } , { ˆ1 } , { , ˆ1 } , { } , { , } , { ˆ1 , } , { } . For n = 8 , there are several possible olutions, each of them consisting of the seven sets { } , { ˆ1 } , { , ˆ1 } , { } , { , } , { ˆ1 , } , { } and one of the sets { , ˆ1 , } , { , } , { ˆ1 , } , { } , respectively. This problem is the enumeration variant (see the recent survey of Epp-stein [18] on enumeration variants) of K th Largest Subset [30] which wasshown to be P -complete by Lawler in 1972 [35]. The procedure given byLawler is designed to be applicable to many different problems and thereforeis not optimal in terms of the running time. To the best of our knowledge, itwas not improved further. Since the applicability of Lawlers procedure to K-Smallest Subsets can be easily missed, we give a specialized algorithm withthe same running time if m equals K and a better running time in every othercase . Theorem 2.
The problem
K-Smallest Subsets is solvable in O (min { K, m } · ( Km + K )) time.Proof. Since P ( A ) contains exponentially many subsets of A , we cannot sim-ply sort them by their sizes if we ask for a polynomial time algorithm. In-stead, we proceed in an iterative way as described in Algorithm 1. We as-sume that the elements of A are sorted in ascending order by their sizes, i.e., s ( a ) ≤ s ( a ) ≤ · · · ≤ s ( a m ) .The idea of the algorithm is the following. Assume we have an array A con-taining the (up to) K smallest subsets of the set A i := { a , . . . , a i } , for some i ∈ { , , . . . , m − } ( A being the empty set), sorted ascendingly by their size.Then we can extend this solution to a solution to K-Smallest Subsets forthe set A i +1 = { a , . . . , a i , a i +1 } : We generate an array B containing the samesubsets as A (in the same order). We then adjoin the element a i +1 to each ofthe subsets in B . The solution to K-Smallest Subsets for A i +1 consists ofthe (up to) K subsets contained in A and B with the smallest size. For findingthese we can make use of the sorting of A and B , and do a two-way merge [32,p. 158] until we find the first (up to) K subsets.We start this procedure with the solution for A , which consists of the emptyset only. Then we extend the solution in an iterative way as described aboveuntil we obtain the solution for A m = A . Correctness.
We show that the following statement I is an invariant for theloop in line (a) of Algorithm 1: I : Array A contains the (up to) K subsets of A i = { a , . . . , a i } withthe smallest size in ascending order.When the algorithm first enters line (a), A contains the only subset of A ,namely ∅ , so I holds. Now we show that I still holds after each repetition of Lawler’s procedure has a running time of O ( Kmc ( m )) , with c ( m ) being the time requiredto compute an optimal solution with m variables. In our case c ( m ) is a constant. Takinginto account the time required to find the minimum in the list [35, Step 1] and the creationof the instances [35, Step 3], this leads to a running time of O ( K ( K + m )) for K-SmallestSubsets . i ≥ of the loop, A stores the (up to) K smallest subsets ofthe last iteration i − . We create (up to) K + 1 smallest subsets of A i which allcontain the element a i , by subjoining a i to each of the subsets that we copy from A , and by creating the set { a i } (line (c)). There cannot exist any a i -containingsubset X of A i which is smaller than the most expensive one in B , because by I , the set X \ { a i } has to be contained in A . By choosing the (up to) K smallestsubsets of A and B to be contained in A for the next iteration, I holds againat the end of the repetition.After min { K, m }− iterations of the outer for-loop in line (a), A contains the so-lution to K-Smallest Subsets . For K ≥ m , this is clear because A m = A . For K < m , the elements a n +1 , . . . , a m cannot be contained in any subset belong-ing to a solution to K-Smallest Subsets , i.e., the solutions to
K-SmallestSubsets for A and A n are identical. Running time.
Most of the operations in Algorithm 1 can be performed inconstant time. The for-loop in line (b) needs to rewrite at most each entry of B . Since B can be represented by a 2-dimensional boolean array with dimension ( K + 1) × m , this can be done in O (( K + 1) · m ) time.In line (e) we compare the sums of the elements of two different subsets. Thesum of the elements of each subset can be stored and maintained (line (c))and can therefore be obtained in constant time for line (e). So, the for-loop inline (d) can be processed in time O ( K ) . The copying process in line (f) takestime O ( K · m ) because of the size of C , but one can avoid this by swapping theroles of A and C in each iteration.We hence obtain a running time of O (min { K, m } · ( Km + K )) .We are now ready to prove some of the open questions that were stated byMattei et al. [41]. Theorem 3. ( OP, A, C any ) -Bribery is in P if A ∈ { IV, DV, IV + DV } .Proof. In this proof, for ease of presentation, the briber’s preferred candidateis called c . Moreover, for every voter v i , we denote by costs i ( c j ) the costs tobribe v i such that c j becomes his favorite candidate. We show Theorem 3 byconstruction of a flow network that can be solved with a minimum cost flowalgorithm (which also maximizes the flow in polynomial time, cf. Ahuja etal. [1]), similarly as initially proposed by Faliszewski [19]. This method hasproven to be useful for showing tractability for many voting problems and isalso used by Mattei et al. [41]. For each r ∈ { , . . . , n } , we check if the voterscan be bribed with budget β such that the preferred candidate c wins with ascore of r votes. If this is possible for at least one r , then we accept, otherwisewe reject. For each ≤ r ≤ n , we define a flow network consisting of a sourcenode s , a target node t , and the following sets of nodes and directed edges. Alledges have costs and capacity , unless specified otherwise. V(oters):
For every voter v i we create a node v i and an edge ( s, v i ) . The setof nodes constructed in this step is called V .14 nput : set A = { a , . . . , a m } with s ( a i ) ∈ Z + , K ∈ N , K ≤ m Output : array A containing the K smallest subsets of A % assumption: s ( a ) ≤ s ( a ) ≤ · · · ≤ s ( a m ) initialize A as an empty array of sets A [1] := ∅ % now A contains only one subset, the empty set (a) for i := 1 , . . . , min { m, K } do initialize B, C as empty arrays of sets % create up to K + 1 new subsets, all containing element a i % ‘ | A | ’ denotes the number of subsets (elements) in array A (b) for j := 1 , . . . , | A | do (c) B [ j ] := A [ j ] ∪ { a i } % find up to K elements with smallest size (like atwo-way merge) indexA , indexB := 1 (d) for j := 1 , . . . , min { K, | A | + | B |} do % by ‘size of A [ k ] ’ we mean the size of the subsetstored in A [ k ] (e) if indexA ≤ | A | and size of A [ indexA ] < size of B [ indexB ] then C [ j ] := A [ indexA ] indexA ++ else C [ j ] := B [ indexB ] indexB ++ (f) A := C return A Algorithm 1:
Pseudocode for solving
K-Smallest Subsets . B(ribery):
For each voter v i , we determine the set consisting of the n cheapestcandidates (the candidates for which the briber has to pay least if hebribes v i to make them his new top candidate) and add c to this set (ifnot present within those n candidates). For each candidate c j of this set,we create a node c ij . Note that at least one candidate with zero costs iscontained in this set (for instance the top candidate). For each such node c ij we create an edge ( v i , c ij ) with costs costs i ( c j ) . The flow on these edgestells the briber who and how he has to bribe. We call the set of all nodescreated in this step B . C’(ollection):
We create a node c ∗ j if there exists a j ∈ [1 , m ] with c ij ∈ B, i ∈ [1 , n ] . For each node c ij ∈ B , we add the edge ( c ij , c ∗ j ) . We call the set ofall nodes created in this step C ′ . Gadget node:
We create one gadget node g and add the edges ( c ∗ j , g ) for allnodes c ∗ j ∈ C ′ \ { c ∗ } . These edges all have capacity r . Finally we add theedge ( c ∗ , t ) with capacity r and the edge ( g, t ) with capacity n − r .15n example for a small instance consisting of two voters is given in Figure 2. s v v c c c c c c c c c ∗ c ∗ c ∗ c ∗ c ∗ g tV B C ′ rrrrrr n − r Figure 2: Example for a flow network of an election with two voters v and v .All edges have costs and capacity . The only exceptions are the dashed edgesthat have a different capacity stated, and the connecting edges ( v i , c ij ) between V and B that have the costs to bribe v i to vote for c j .As explained above, we create n such networks, each differing only in thevalue of r which is the maximum score all candidates can reach and the preferredcandidate c will win with, and solve the corresponding flow problem with aminimum cost flow algorithm. The resulting flow will always be n , but the costsmay differ. We then choose the solution with the lowest costs, and if they donot exceed the given budget, we found an optimal one. Otherwise there is none.The constructed networks have a number of nodes that is polynomial in n andwe are only building up to n of them. This results in a polynomial running timeof the overall algorithm.For the correctness, observe that there is a flow of value n on the constructednetwork if and only if the bribery problem can be solved: Assume there existsa flow of value n in the network. Since all capacities and costs are integral,the flow is integral as well [1]. The edges connecting nodes in C ′ with g and t ,respectively, guarantee that a flow of value r is going through vertex c ∗ , henceensuring that the preferred candidate c obtains r points, and that no flow ofvalue greater than r leaves any vertex c ∗ j ∈ C ′ \ { c ∗ } , hence ensuring that nocandidate can end up with more than r points, making c a (co-)winner. This isachieved if the briber changes those votes corresponding to the edges connecting V and B : For each node v i , there is one edge ( v i , c ij ) connecting v i to a node c ij of the set B , carrying a flow of value 1. This edge tells the briber that v i hasto be bribed in such a way that c j becomes his new top candidate. Therefore,there exists a successful bribery in which c wins with r votes.Conversely, given a bribery that makes c a winner of the election withoutexceeding budget β , we can construct a flow in the following way: Since c is16 winner of the bribed election, he must have obtained a total of r points forsome ≤ r ≤ n . So we choose the network with capacity r on the edges ( c ∗ j , g ) and ( c ∗ , t ) . We set the flow on the edges ( s, v i ) with v i ∈ V to 1. We distinguishtwo types of voters: If after the bribery, candidate c j is on the top positionfor voter v i and the node c ij is created for B (meaning that c j is among thecheapest n candidates for v i ), we set the flow on the edges ( v i , c ij ) and ( c ij , c ∗ j ) to 1, otherwise we call the voter v i an overpaid voter and ignore him for now.Note that c i is a node in B for each i . Having set as much flow on the edgesbetween nodes of V and C ′ as this rule allows, we carry on preserving the flowfrom nodes in C ′ to t . This is easy, because there is exactly one path from eachnode in C ′ to t . For each voter v k in the set of overpaid voters, the flow on theedges between the node v k and nodes of B as well as on the edges leaving thesenodes of B has not been set yet. For each such voter v k , we just have to find an augmenting path (cf. the Edmond-Karp algorithm [15] for the Ford-Fulkersonalgorithm [29] for solving the flow problem) from v k to t . There are at least n nodes in C ′ and n different nodes c kj for each v k , so by the pigeon hole principle,there exists at least one such path, and the cost of each such path is lower thanthe amount paid by the briber to bribe v k . Therefore there is a flow of value n for this network with costs of at most β .There are two more things to be observed. First, as described by Mattei etal. [41, Theorem 8], for each voter, it suffices to choose the n cheapest candidateswhen constructing the set B in the network instead of including all m possiblecandidates: There are n − other voters who can give one vote each to acandidate. By the pigeonhole principle, it is not possible that each of the n cheapest candidates for a voter already has a score of r votes, hence in anoptimal solution, no additional (more expensive) candidate has to be bribed.Second, the selection of the nodes in B corresponds to solving the K-SmallestSubsets problem, where the sizes of the elements in A correspond to the allowedindividual costs for flipping every issue of voter v i . This can be done with thealgorithm presented in the proof for Theorem 2. Each subset F belonging tothe solution to the K-Smallest Subsets algorithm represents one candidate,where the size of the elements in F are the costs of the required flips to bribe v i to vote for this candidate. To cope with the different bribery actions, one cansimply adjust A . There are no further changes needed.Theorem 3 implies that ( OP, A, C level ) -bribery can be solved in polynomialtime as well, because C level is a special case of C any . In fact, the same holdsfor the cost schemes C flip and C dist , but the complexity for those was alreadyshown by Mattei et al. [41]. Corollary 4. ( OP, A, C level ) -Bribery is in P if A ∈ { IV, DV, IV + DV } . Theorem 3 can be extended to the voting system OK for O -legal profiles,when k is a power of , which is called OK ∗ then. To show this, we use Lemma from Mattei et al. : Lemma 5 (Mattei et al. [41]) . Given an acyclic CP-net X and a constantlinearization scheme across all agents and k = 2 j , for some j ∈ N , the top k utcomes of X are all the outcomes differing from the top one on the value ofexactly j issues. Moreover, given two CP-nets in the same profile and with thesame top element, they have the same top k elements. We recall that existence of a constant linearization scheme across all agentsmeans O -legality of the profile for some given strict total order O . Lemma implies that any ( OK ∗ , A, C ) -Bribery instance can be treated as a ( OP, A, C ) -Bribery instance, for any cost scheme C and bribery action A , by ignoring thelast log k issues. Corollary 6. ( OK ∗ , A, C ) -Bribery with bribery action A ∈ { IV, DV, IV + DV } and cost scheme C ∈ { C any , C level } is in P . Similarly as in the proof of Theorem 3, one can show tractability for thevoting system OV with cost scheme C any . Theorem 7. ( OV, A, C any ) -Bribery is in P if A ∈ { IV, DV, IV + DV } .Proof. We distinguish two cases: n < m and n ≥ m . The first case is straight-forward, the second case can be translated into a network flow problem. Thesolution flow indicates who has to be bribed and how such that the preferredcandidate p wins the election. n < m : By the pigeonhole principle, there must exist at least one candidatewho got no veto at all and is therefore the winner of the election. Thepreferred candidate p has to get rid of all his vetoes to be one of thewinners. Therefore the briber has to bribe every voter who vetoes against p , by choosing the cheapest bribery possible. n ≥ m : We construct a flow network similar to the one in the proof of The-orem 3. Again, the preferred candidate p of the briber is called c . Wetake a source s and a target t for the network and construct the followingsets of vertices and edges. All edges have costs and capacity , unlessspecified otherwise. General:V(oters):
For every voter v i we create a node v i and an edge ( s, v i ) .These edges each have capacity m − . A(pproval):
For each voter v i and each candidate c j who gets approvedby v i , we create a node ˆ c ij together with the edge ( v i , ˆ c ij ) . B(ribery):
For each voter v i and every candidate c i , we create a node c ij .Let c x be the candidate v i vetoes against. For each candidate c j = c x we then create the edge (ˆ c ij , c ij ) with costs . If it is allowed (depend-ing on the bribery action) to bribe voter v i to veto for candidate c j ,we additionally create the edge (ˆ c ij , c ix ) with the costs to bribe v i thisway. Flow in the solution on those additional edges indicates the vetois bribed from c x to c j . Note that the costs for the edge (ˆ c ij , c ix ) canbe calculated in time polynomial in m .18 ’(ollection): For each candidate c j we create a node c ∗ j . For each node c ij created in the previous step, we add the edge ( c ij , c ∗ j ) . Gadget-node:
We create one gadget-node g and add the edges ( c ∗ j , g ) forall nodes c ∗ j created in the previous step except for c ∗ . These edgesall have capacity r . Finally we add the edge ( c ∗ , t ) with capacity r and the edge ( g, t ) with capacity n · (2 m − − r .An example is given in Figure 3. s v v v ˆ c ˆ c ˆ c ˆ c ˆ c ˆ c c c c c c c c c c c c c g tV A B C ′ . . . δδδ rrrr γ Figure 3: Example of a flow network created from an election with three votersand m = 2 resulting in candidates. Only the first two voters are shown indetail. Voter v casts his veto against c , while voter v casts his veto against c . The costs of all edges are except for the dashed (ˆ c ij , c ix ) ones which carrythe costs to bribe v i in a way to veto against c j (instead of c x ). The capacitiesare all except for the edges where a different value is specified in the graph.Here, δ = 2 m − and γ = n · (2 m − − r , where r ranges from 1 to n for the n networks constructed.We proceed as in the proof of Theorem 3. We again construct a flow networkfor each r with ≤ r ≤ n · (2 m − and take the solution with a flow of n · (2 m − and minimal costs as the solution to the bribery problem. Again, solving theflow problem is equivalent to solving the bribery problem. We remark that thecase n = 2 m can also be handled with a smaller network.As before, C level is a special case of C any , implying the following corollary. Corollary 8. ( OV, A, C level ) -Bribery is in P if A ∈ { IV, DV, IV + DV } . .2 Results for the weighted case In this section, we analyze the weighted version of the bribery problem. Mattei et al. [41] also consider the case of weighted voters for the voting system SM forwhich they obtain N P -completeness. To the best of our knowledge, no resultin the weighted case for OP , OV , and OK has been published yet. We show foralmost all of these remaining systems that the bribery problem is N P -completeas well by reductions from the
Partition problem.The idea in all of the reductions in this section is the following. We constructan election in which one candidate called c is the winner with a total of ψ points which he obtains by a set of voters whose weights correspond to theelements in the set to be partitioned. The preferred candidate p obtains ψ points. The instance is built in such a way that for the briber, the only way tomake p win consists in transferring points from c to another candidate c , thussolving the Partition problem, such that c , c and p end up with the samescore of ψ as co-winners. Theorem 9.
Weighted- ( OP, A, C ) -Bribery is N P -complete with bribery ac-tion A ∈ { IV, DV } and cost scheme C ∈ { C equal , C any , C level , C dist , C flip } .Proof. We show Theorem 9 by reduction from
Partition . Let I P be an in-stance of Partition with A = { a , . . . , a ℓ } and P a ∈A a = 2 ψ . The instance I B of Weighted- ( OP, A, C ) -Bribery that is constructed from I P has two is-sues X and Y with domains { x, x } and { y, y } , respectively, and therefore fourcandidates. For the bribery action DV these are the following: xy The preferred candidate p of the briber. He has one ψ -weighted voter votingfor him. xy The candidate c , who starts as the winner with ψ votes. xy The candidate c , for whom no one votes initially, but who will win togetherwith p and c if there exists a partition. xy The candidate c who is not important for the case of bribery action DV but for the case of bribery action IV .We achieve this by constructing the following voters:voter top candidate dependency weight voting for v xy none ψ pv xy X → Y a c v xy X → Y a c ... v ℓ +1 xy X → Y a ℓ c Y for everyvoter except v , so he cannot transfer points from c to p . He can just split thepoints of c among c and c , in which case they win together with p . In everyother case p loses.More formally, we show that I P is a yes-instance if and only if I B is a yes-instance. Let A ′ ⊂ A be a solution to I P . Then the briber bribes those voterswhose weights are the elements of A ′ . This results in a score of ψ for p, c and c , so I B is a yes-instance.Conversely, let I B be a yes-instance, implying there is a bribery that makes p a(co-)winner of the given election. Since the briber can only bribe the dependentissue Y , he can make the voters v , . . . , v ℓ +1 vote for the candidate xy = c only.The only possible way to make p a (co-)winner is hence to split the points of c among c and c so that they both end up with a score of ψ . The weights of thevoters that have to be bribed to make them vote for c are then the elements ofthe subset A ′ ⊂ A that solves I P .Since no costs are involved in this reduction, it works for every cost schemeconsidered, assuming an unlimited budget.For the case IV , we can use the same construction. In this case, the bribercan only bribe the independent issue X , hence split the points of c among c and c = xy , which is again the only possible way to make p a winner.Since OP is a special case of OK , we obtain the following corollary. Corollary 10.
Weighted- ( OK, A, C ) -Bribery is N P -complete for briberyaction A ∈ { IV, DV } and cost scheme C ∈ { C equal , C any , C level , C dist , C flip } . For the voting rule OV we can also reduce from the partition problem whenwe restrict the briber to the DV actions only and obtain the following theorem.The cases of IV and IV + DV are not covered yet. Theorem 11.
Weighted- ( OV, DV, C ) -Bribery is N P -complete for each costscheme C ∈ { C equal , C any , C level , C dist , C flip } .Proof. We show Theorem 11 by reduction from
Partition once more. Let I P be an instance of Partition with A = { a , . . . , a ℓ } and P a ∈A a = 2 ψ . Theinstance I B of Weighted- ( OV, DV, C ) -Bribery that is constructed from I P has two issues X and Y with domains { x, x } and { y, y } , respectively, andtherefore four candidates. For the bribery action DV these are the following: xy The preferred candidate p of the briber. There is one voter with weight ψ casting his veto against him. xy The candidate c , who initially receives ψ vetos. xy The candidate c , against whom no one casts a veto initially, but who willwin together with p and c if there exists a partition. xy An unimportant clone u of p . 21e achieve this by constructing the following voters:voter top candidate dependency weight vetoes agains v xy none ψ pv xy none ψ uv xy X → Y a c v xy X → Y a c ... v ℓ +2 xy X → Y a ℓ c Note that the briber is only allowed to bribe the dependent issue Y for everyvoter except v , v , so he can only transfer vetos from c to c . The best he cando is to split the vetos of c among c and c , in which case they win togetherwith p and u . In every other case p loses. As in the proof of Theorem 9, the twoinstances are equivalent. As no costs are involved here, the reduction works forevery cost scheme considered, assuming an unlimited budget.We now investigate the weighted case for the voting rule SM . We found thatthe situation here has to be analyzed in more detail: Hardness of the weightedversion of the bribery problem can be caused just by the use of the cost vector.Mattei et al. [41, Theorem 6] showed that Weighted- ( SM, A, C ) -Bribery is N P -complete for all bribery actions and for all cost schemes they considered.However, for the cost schemes C ∈ { C level , C dist , C flip , C any } , their reductionimplicitly makes use of the cost vector Q . Hence, the computational hardnessoriginates from the individual costs for each voter.This is not the case for C equal , where N P -completeness already follows fromthe result in the unweighted case, without any requirement on individual costs.Mattei et al. observe that for the cost scheme C flip , if the cost vector Q con-tains only ones, Weighted- ( SM, A, C flip ) -Bribery is in P with bribery action A ∈ { IV, DV, IV + DV } ([41, Theorem 7]). We found the same property for Weighted- ( SM, A, C level ) -Bribery , which is also solvable in polynomial timefor all bribery actions considered in this work if the cost vector contains onlyones. Theorem 12.
Weighted- ( SM, A, C level ) -Bribery with bribery action A ∈{ IV, DV, IV + DV } is, for the special case that Q [ i ] = 1 for each voter v i ,solvable in time O ( n m ) , .Proof. For the voting system SM , it is sufficient to look at the issues one afteranother, since they do not affect each other, even with dependencies. For eachissue, a certain amount of weighted voters has to be bribed to flip the value ofthe corresponding issue. The briber just needs to know which of the voters arethe cheapest ones to bribe. We first start with bribery action IV + DV .Let V be the set of all voters, let w i be the weight of voter v i , W = P ni =1 w i the sum of all weights, and costs ( v i ) the cost to bribe voter v i in the consideredissue. The set of all voters who vote in the given issue for the same value22hat is taken by p in this issue is called G for the good voters, and B = V \ G denotes the set of the remaining beneficiary voters. The briber is interested infinding a subset S ′ ⊆ B , such that P v i ∈S ′ ∪G w i > ⌊ W ⌋ and P v i ∈S ′ costs ( v i ) isminimal. This is the same as finding a subset S ⊆ B , such that P v i ∈S w i ≤ ⌊ W ⌋ and P v i ∈S costs ( v i ) is maximal. The latter problem is the same as solving the Knapsack problem on the set B where the bribery costs costs ( v i ) correspond tothe values, and the weights of the voters correspond to the size of the items to bepacked in the knapsack, respectively. The crucial observation here is that with C level , the cost for flipping the value of the relevant cp-statement is bounded bythe number m of issues, as there cannot be more levels than issues by definition,hence costs ( v i ) ≤ m for all v i ∈ V . As mentioned in Section 2, one can solvethe Knapsack problem with dynamic programming in time O ( n · T ) , where T is the sum of the values of all objects. In our case, T is the sum P ni =1 costs ( v i ) ,hence bounded by n · m , so that the running time of the dynamic programmingalgorithm proposed by Dantzig (see [12]) is polynomial. In detail, we implementDantzig’s solution as follows:We define an n × P v i ∈B costs ( v i ) matrix D with entries d i,j being the min-imum weight of all subsets S ⊆ { v , . . . , v i } with P v i ∈S costs ( v i ) = j . Setting d i, = 0 for all i , d − ,j = ∞ for j > , and d i,j = ∞ for j < , we can computeall entries of D recursively by d i,j = min ( d i − ,j ,d i − ,j − costs ( v i ) + w j , with w i and costs ( v i ) being the weight of voter v i and costs to bribe him in theconsidered issue, respectively. The optimal solution can then be found as theentry d |B| ,j with maximum j such that d |B| ,j ≤ ⌊ W ⌋ . The set of voters (whichwe do not bribe) corresponding to this value can be found by backtracking.For the considered issue, we hence have a matrix with O ( n · nm ) entries,each of which can be calculated in O (1) time. This has do be done for eachissue, so the algorithm has an overall running time of O ( n m ) .With bribery actions IV and DV , the briber is not able to bribe every voterfor specific issues. Therefore we partition the set of beneficiary voters B inthe set of unbribable voters B U and the set of bribable voters B B . We are thensearching for a set S ⊆ B B with P v i ∈S costs ( v i ) maximal such that P v i ∈S w i ≤⌊ W ⌋ − P v j ∈B U w j . Such a set can be obtained with the same algorithm asdescribed above. In this section, we analyze the complexity of the unweighted negative briberyproblem in CP-nets. For the voting system SM combined with cost scheme C equal , we can easily adapt the proof of the unweighted positive case and ob-tain N P -completeness as well. Since we use a (parameterized) reduction from
Negative Optimal Lobbying , this also means that the problem is W [2] -hardwith respect to the budget, cf. the remark in Section 2.3.23 heorem 13. ( SM, A, C equal ) -negative-Bribery with bribery action A ∈{ IV, DV, IV + DV } is N P -complete.Proof.
Theorem 13 can be shown by modifying the proof for ( SM, A, C equal ) -Bribery being N P -complete [41, Theorem 5]. We just need to exchange theterm ‘
Optimal Lobbying ’ by ‘
Negative Optimal Lobbying ’. Theorem 14. ( SM, A, C ) -negative-Bribery is in P for a bribery action A ∈{ IV, DV, IV + DV } and a cost scheme C ∈ { C any , C level , C flip } . s v v v X free X X X X X v v tP F DE Figure 4: Example for the flow network of step 2 in the proof of Theorem 14to adjust a solution for ( SM, A, C ) -Bribery to the negative version. In thesolution to the non-negative version, the voters v , v , and v are voting for p = 111 , v for , and v for . All edges have costs and capacity . Theonly two exceptions are the dashed ones going to target t , and the dotted onesbetween P and E . The former ones have a different capacity stated, and thelatter ones have different costs. The costs of the edges ( v i , X jl ) and ( v i , X free l ) depend on whether v i was bribed in issue X l . Proof.
We construct the solution in two steps. First, we apply the polynomialtime algorithm given by Mattei et al. [41, Theorem 4] to solve an ( SM, A, C ) -Bribery instance. Here, a greedy strategy helps to choose which voter is to bribein which issue. Unfortunately, the solution may contain some voters who arebribed to vote directly for p , therefore it does not solve ( SM, A, C ) -negative-Bribery directly. In the second step we repair the first solution by choosing thecheapest flips such that no voter votes for p . This is done with a flow network.As the number of nodes in this network polynomially depends on the number ofvoters and issues, the flow problem on the network can be solved in polynomialtime, too. 24n example is given in Figure 4.We take a source s and a target t for the network and construct the followingsets of vertices and edges. All edges have costs and capacity , unless specifiedotherwise. P(-Voters):
For every voter v i voting for p we create a node v i and an edge ( s, v i ) . D(onor Voters):
For each voter v j not voting for p we create a node v j andan edge ( v j , t ) . Let d be the number of issues in which v j ’s top candidatediffers from p . Then the capacity of the edge ( v j , t ) is set to d − . Thisensures that v j cannot vote for p . F(ree Issues):
For each issue X l , in which k > ⌊ n ⌋ + 1 voters vote for p , wecreate a node X free l and an edge ( X free l , t ) , with capacity k − ( ⌊ n ⌋ + 1) .Additionally, we create an edge ( v i , X free l ) for each v i created in P withcosts costs i ( X l ) . Flow on these edges indicates which voter has to bebribed in which issue. E(xpensive Issues):
For each node v j in D we create a node X jl if voter v j is not voting for p in issue X l . For each of these nodes we create anedge ( X jl , v j ) . Additionally, we create all possible edges between eachnode v i in P and each node X jl in E . Those edges get different costsdepending on whether v i was bribed in issue X l . If so, they get the cost costs j ( X l ) − costs i ( X l ) , and costs j ( X l ) + costs i ( X l ) , otherwise. Again,flow on these edges indicates which voter has to be bribed in which issue.As in the proof of Theorem 3, one can show that there exists a flow of value n if and only if the corresponding bribery problem can be solved. The idea hereis that at least one issue has to be flipped for each voter voting for p . Thiscan be done easily if more than ⌊ n ⌋ + 1 voters are voting for p in this issue.Otherwise the same issue has to be flipped in the vote of a voter who is notvoting for p . Additionally, one has to prohibit those voters to vote for p in theend. This is ensured by the capacities on the edges to the target t . This way,the flow algorithm can find the cheapest way to repair the solution of the firststep, which leads to an optimal solution.For OP , we can adapt the proof of the non-negative case (Theorem 3) andobtain the following theorem. Theorem 15. ( OP, A, C ) -negative-Bribery is in P with bribery action A ∈{ IV, DV, IV + DV } and a cost scheme C ∈ { C equal , C flip , C level , C any , C dist } .Proof. We can almost adopt the proof for Theorem 3 here, we just have tochange the definition of the set of nodes B : Whenever p is not the top candidateof a voter v i , we do not create the node c i and its connecting edges. Then, avoter cannot be bribed to vote for p . This works for C any , C level , C flip , and C dist . For C equal , we need additional changes. Since all changes have the same25osts, we can choose n +1 arbitrary candidates ( = p ) for each voter to create thenodes in B together with the one for the top candidate, instead of using onlythe cheapest ones as before. Finally, the costs for the edges ( v i , c ij ) connecting V with B are set to if c j is the top candidate of v i , or otherwise.To prove the next theorem, we need Lemma from Mattei et al. [41] once more. Theorem 16. ( OK ∗ , A, C ) -negative-Bribery is in P with bribery action A ∈{ IV, DV, IV + DV } and cost scheme C ∈ { C equal , C flip , C level , C any , C dist } .Proof. Due to Lemma 5 [41] we know that for OK ∗ , we only need to considerthe top candidate instead of the first k candidates, or, to be more precise, justthe values of the first m − j issues of the top candidate. So we just need to solvean instance of ( OP, A, C ) -negative-Bribery which is possible in polynomialtime due to Theorem 15.Adapting the proof of Theorem 7, we obtain the following theorem for thevoting system OV . Theorem 17. ( OV, A, C ) -negative-Bribery is in P with bribery action A ∈{ IV, DV, IV + DV } and cost scheme C ∈ { C equal , C flip , C level , C any , C dist } .Proof. We can almost adapt the proof for Theorem 7. We once more distinguishthe two cases n < m : If p got at least one veto, it is a No-instance. Otherwise it is a Yes-instance. n ≥ m : In this case we create one flow network for each r with ≤ r ≤ n · (2 m − as described in the proof of Theorem 7. However, this time,we do not create the node c i and its incident edges if a voter v i is castinghis veto against c . The rest remains unchanged. In this subsection, we derive hardness results for all variants of the weightednegative case of the bribery problem, again by reductions from the
Partition problem following the main idea described in Subsection 3.2.We have shown in Theorem 9 in Subsection 3.2 that the weighted version ofthe bribery problem is
N P -complete for OP for all cost schemes and briberyactions IV and DV . In the given proof, the briber is not able to transfer anypoints to his preferred candidate p directly, hence this also solves the negativecases for bribery actions IV and DV . For the negative case, we can extend thisresult to bribery action IV + DV by adding a voter v ℓ +2 with weight ψ votingfor c without dependencies. Then p can only win if the briber is once moreable to split the votes for c between c and c . We hence obtain the followingcorollary for Theorem 9. 26 orollary 18. Weighted- ( D, A, C ) -negative-Bribery is N P -complete fora voting rule D ∈ { OP, OK } , a bribery action A ∈ { IV, DV, IV + DV } , and acost scheme C ∈ { C equal , C any , C level , C dist , C flip } . To show
N P -completeness for the voting rule OV , we can use the reductiongiven in the proof of Theorem 11 in Subsection 3.2 for the positive version withbribery action DV , since in this proof, the briber never asks a voter to cast hisveto from p to another candidate. We will provide yet another reduction from Partition to cover IV and IV + DV in the negative case, too. Summarizing,we obtain the following theorem. Theorem 19.
Weighted- ( OV, A, C ) -negative-Bribery with bribery action A ∈ { IV, DV, IV + DV } and cost scheme C ∈ { C equal , C any , C level , C dist , C flip } is N P -complete.Proof.
Theorem 19 follows directly from Theorem 11 for the bribery action DV .For the bribery actions IV and IV + DV , we use a similar reduction as inthe proof of Theorem 11. Let I P be an instance of Partition with A = { a , . . . , a ℓ } and P a ∈A a = 2 ψ . The instance I B of Weighted- ( OV, A, C ) -negative-Bribery that is constructed from I P has two issues X and Y withdomains { x, x } and { y, y } , respectively, and therefore the following four candi-dates: xy The preferred candidate p of the briber. There is one voter with weight ψ casting his veto against him. xy The candidate c , who starts with a weighted sum of · ψ vetos. xy The candidate c , against whom no one casts a veto initially, but who willwin together with the other 3 candidates, if there exists a partition. xy An unimportant clone of p .We achieve this by constructing the following voters:voter top candidate dependency weight vetoes against v xy none ψ pv xy none ψ uv xy X → Y a c v xy X → Y a c ... v ℓ +2 xy X → Y a ℓ c Note that due to the bribery action IV , the briber is only allowed to bribe theindependent issue X for every voter except v , v , so he can only transfer theirvetos from c to c . He is not allowed to bribe voter v in the case of negativebribery, and it is of no use to him to bribe voter v . The best he can do is tosplit the vetoes of c among c and c , in which case they win together with p u . In every other case p loses. With the same argument as in the proof ofTheorem 9, the two instances are equivalent. Since no costs are involved here,this works for every cost scheme considered, assuming an unlimited budget.The same holds for the bribery action IV + DV . Here, the briber could bribethe voters v i , i ≥ to cast their veto against c , c , and u , but this does nothelp because the candidate with the fewest vetos wins and there is a total sumof ψ vetos to distribute among u, c , and c .Last, we show that Weighted- ( SM, A, C ) -negative-Bribery is compu-tationally hard, independent of the cost scheme that is used. Theorem 20.
Weighted- ( SM, A, C ) -negative-Bribery with a bribery ac-tion A ∈ { IV, DV, IV + DV } and any cost scheme C is N P -complete.
The proof uses a reduction from
Partition where no costs are involved.
Proof.
We prove Theorem 20 by reduction from
Partition . Let I P be aninstance of Partition with A = { a , . . . , a ℓ } and P a ∈A a = 2 ψ . The instance I B of Weighted- ( SM, A, C ) -negative-Bribery that is constructed from I P has three issues X, Y, Z with domains { x, x } , { y, y } , { z, z } , respectively, andtherefore eight candidates. For the bribery action DV , only the following fourof them are important: xyz The preferred candidate p of the briber. There is one voter voting for himweighted with . xyz The candidate c , for whom no one votes initially. xyz The candidate c , for whom no one votes initially, neither. The two candi-dates c and c serve as the two partitions. xyz The candidate c , for whom almost everyone votes initially, but who willhave no voter voting for him if p wins.We achieve this by constructing the following voters:voter top candidate dependency weight voting for v xyz none pv xyz X → Y, X → Z a c v xyz X → Y, X → Z a c ... v ℓ +1 xyz X → Y, X → Z a ℓ c Note that the briber is only allowed to bribe the dependent issues Y or Z forevery voter except v , so he cannot transfer points from c to p . He can justsplit the points of c equally among c and c , in which case p wins. In everyother case p loses. As in the proof of Theorem 9, one can show that the twoinstances are equivalent. As no costs are involved here, this works for every cost28cheme considered, assuming an unlimited budget.For the case IV we use a slightly different construction. In this case we donot need the issue X , which makes the issues Y and Z independent with thereduction still working. The latter version has only independent issues, so thisreduction works as well for the bribery action IV + DV . Our results are summarized in Table 2.The bribery problem and the variants of microbribery, nonuniform bribery,and swap bribery in the ‘classical’ setting of unconditional preferences given aslinear orders are tractable for the voting systems plurality and veto ([21], [23],[19], [17]). One might expect that bribing turns out to be more difficult in thecase of conditional preferences and as soon as more complex cost schemes areused, but it does not. For the non-negative unweighted version of the briberyproblem, Mattei et al. [41] obtained several tractability results (see Table 2).We could solve the remaining unknown complexities for the cost schemes C any and C level . Contrary to the conjecture of Mattei et al. [41], it turned out thatthe bribery problem is easy in these cases as well.These easiness results could be explained by the fact that OP and OV onlyrequire very little information on the voters’ (conditional) preferences. But,more importantly, using a one-step voting rule, the bribery of one CP-net doesnot have an influence on other CP-nets. This is different for the sequentialvoting rule SM . If the value of an issue is changed due to bribery of one CP-net, the cp-statements of dependent issues in other CP-nets are concerned aswell, and this—in combination with the cost scheme—can make the problempotentially hard.The interesting case might be the voting rule OK : In the classical setting, thebribery problem is polynomially solvable for k -approval elections [21], whereas Swap Bribery is N P -complete for k ≥ [3, 17]. So far, for bribery in CP-nets, only results for the special case of OK for O -legal profiles and where k isa power j of 2 (denoted by OK ∗ by Mattei et al. ) are known; it was shown byMattei et al. [41] that the bribery problem is solvable in polynomial time then.This is due to the fact that in those cases, a voter always approves one package of k candidates out of m − j such packages, which are all fixed and disjoint. It isso to speak just a slightly different version of OP . It would hence be interestingto investigate the computational complexity for other values of k .For the non-negative weighted case, we could show that finding an optimalbribery is N P -complete for OP , OV , and OK for all considered cost schemes.However, not all bribery actions are covered yet. For OP , OV and OK , thecomputational hardness is due to the weights which enforce that a partitionproblem has to be solved—this is typical for the weighted variant of a problem,cf. the weighted versions of the original family of bribery problems in the work29f Faliszewski et al. [21]. However, it is interesting to see that weighted vot-ers do not necessarily make the problem hard—this only holds in the negativecase where we could show N P -completeness for all considered variants of theproblem. We have seen that the complexity for SM in the non-negative casedepends on the choice of the cost scheme and the cost vector, not only on theweights.The complexity for SM with cost scheme C dist remains unsolved for the un-weighted cases.Table 2: Complexity results for variants of the bribery problem in CP-nets. Wedistinguish solvability in polynomial time ( P ) and N P -completeness (
N P -c).The given results all hold for the bribery actions IV , DV , and IV + DV , exceptfor the results in the rows marked with † , which are only shown for briberyactions IV and DV so far, and the ones in the row marked with ‡ , which areonly shown for bribery action DV . Results in bold face are obtained in thispaper, the results in light typeface are due to Mattei et al. [41]. OK ∗ is thespecial case of OK when k is a power of 2 and an O -legal profile is given. Theresults marked with ♦ are partly shown by Mattei et al. [41]; they show theresult only for the bribery case IV , in the paper on hand it is shown to hold for IV , DV and IV + DV . The cases labeled with more than one complexity classcan be solved in polynomial time if the cost vector Q contains a for each voter,and are N P -complete for arbitrary cost vectors. In all the remaining tractablecases, the corresponding problem remains in P even for arbitrary cost vectors,while all of our hardness results still hold with Q [ i ] = 1 for all ≤ i ≤ n . C equal C flip C level C any C dist SM N P -c P P P ?OP
P P P ♦ P P Cor.4/Thm.3 OV P P P ♦ P P Cor.8/Thm.7
OK*
P P P ♦ P P Cor.6 weighted SM
N P -c P , N P -c P , N P -c N P -c N P -c Thm.12 OP † N P -c N P -c N P -c N P -c N P -c Thm.9 OV ‡ N P -c N P -c N P -c N P -c N P -c Thm.11 OK † N P -c N P -c N P -c N P -c N P -c Cor.10 negative SM
N P -c P P P ? Thm.13/14 OP P P P P P
Thm.15 OV P P P P P
Thm.17
OK*
P P P P P
Thm.16 weighted SM
N P -c N P -c N P -c N P -c N P -c Thm.20 negative OP
N P -c N P -c N P -c N P -c N P -c Cor.18 OV N P -c N P -c N P -c N P -c N P -c Thm.19 OK N P -c N P -c N P -c N P -c N P -c Cor.18
Summarizing, the unweighted versions do seem particularly appealing forelection campaign management due to their tractability. The only exception is SM , depending on the cost scheme used. The weighted versions of SM are all30 P -complete, with tractability for the cost schemes C flip and C level if weighted SM is used without individual voter costs. Of course, N P -completeness resultsfor the considered problems only constitute a worst case analysis and thereforecannot guarantee resistance against manipulative actions. However, they canhelp in acquiring a better understanding of the structure of the underlying prob-lems and therefore may contribute in finding heuristic approaches to practicallydeal with them, or provide a structural decomposition for further investigationsfrom the point of view of parameterized complexity [42], which again is a desiredproperty in the setting of campaign management.In respect of future research, we hope that our work contributes in creating amore extensive understanding of the nature of voting with CP-nets. The land-scape of complexity in the ‘classical’ setting where voters have unconditionalpreferences given as linear orders over the candidates is already quite elaborate,and it would be interesting to obtain a similar overview of the complexity of dif-ferent voting problems for the CP-net setting as well. This includes the study ofadditional voting rules and other common voting problems. Considering differ-ent voting problems for the setting of CP-nets such as the manipulation problemas initiated by Mattei [39] or election control will be of value for measuring vul-nerability and resistance of voting in CP-nets.Another interesting extension is the case that the dependencies of some issuesare linked to those of other voters, as proposed in the setting of m CP-nets byRossi et al. [47]. In Example 1, Bob might prefer where to go depending onAlice’s choice of the destination, or even on Alice’s preference when to go orwhat to do in the holiday. Mattei et al. [41] have also suggested to allow thebriber to create dependencies instead of only deleting them, and to pay votersto create preferences that are conditioned by those of other voters.
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