     # Frugal Bribery in Voting

Bribery in elections is an important problem in computational social choice theory. However, bribery with money is often illegal in elections. Motivated by this, we introduce the notion of frugal bribery and formulate two new pertinent computational problems which we call Frugal-bribery and Frugal- bribery to capture bribery without money in elections. In the proposed model, the briber is frugal in nature and this is captured by her inability to bribe votes of a certain kind, namely, non-vulnerable votes. In the Frugal-bribery problem, the goal is to make a certain candidate win the election by changing only vulnerable votes. In the Frugal-dollarbribery problem, the vulnerable votes have prices and the goal is to make a certain candidate win the election by changing only vulnerable votes, subject to a budget constraint of the briber. We further formulate two natural variants of the Frugal-dollarbribery problem namely Uniform-frugal-dollarbribery and Nonuniform-frugal-dollarbribery where the prices of the vulnerable votes are, respectively, all the same or different. We study the computational complexity of the above problems for unweighted and weighted elections for several commonly used voting rules. We observe that, even if we have only a small number of candidates, the problems are intractable for all voting rules studied here for weighted elections, with the sole exception of the Frugal-bribery problem for the plurality voting rule. In contrast, we have polynomial time algorithms for the Frugal-bribery problem for plurality, veto, k-approval, k-veto, and plurality with runoff voting rules for unweighted elections. However, the Frugal-dollarbribery problem is intractable for all the voting rules studied here barring the plurality and the veto voting rules for unweighted elections.

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## 1 Introduction

In a typical voting scenario, we have a set of candidates and a set of voters reporting their preferences or votes which are complete rankings of the candidates. A voting rule is a procedure that, given a collection of votes, chooses one candidate as the winner. A set of votes over a set of candidates along with a voting rule is called an election.

Activities that try to influence voter opinions, in favor of specific candidates, are very common during the time that an election is in progress. For example, in a political election, candidates often conduct elaborate campaigns to promote themselves among a general or targeted audience. Similarly, it is not uncommon for people to protest against, or rally for, a national committee or court that is in the process of approving a particular policy. An extreme illustration of this phenomenon is bribery — here, the candidates may create financial incentives to sway the voters. Of course, the process of influencing voters may involve costs even without the bribery aspect; for instance, a typical political campaign or rally entails considerable expenditure.

All situations involving a systematic attempt to influence voters usually have the following aspects: an external agent, a candidate that the agent wishes to win (or lose) the election, a budget constraint, a cost model for a change of vote, and knowledge of the existing election. The formal computational problem that arises from these inputs is the following: is it possible to make a distinguished candidate win the election in question by incurring a cost that is within the budget? This is the problem of bribery from a computational perspective. This question, with origins in [Faliszewski et al., 2006, Faliszewski et al., 2009a, Faliszewski et al., 2009b], has been subsequently studied intensely in computational social choice literature. In particular, bribery has been studied under various cost models, for example, uniform price per vote which is known as $Bribery [Faliszewski et al., 2006], nonuniform price per vote [Faliszewski, 2008], nonuniform price per shift of the distinguished candidate per vote which is called the Shift bribery, nonuniform price per swap of candidates per vote which is called the Swap bribery [Elkind et al., 2009]. The campaigning problem has also been studied for various vote models, for example, truncated ballots [Baumeister et al., 2012], soft constraints [Pini et al., 2013], CP-nets [Dorn and Krüger, 2014], combinatorial domains [Mattei et al., 2012] and probabilistic lobbying [Erdélyi et al., 2009]. The bribery problem has also been studied under voting rule uncertainty [Erdelyi et al., 2014]. Faliszewski et al. [Faliszewski et al., 2014] study the complexity of bribery in Bucklin and Fallback voting rules. Xia [Xia, 2012] studies destructive bribery, where the goal of the briber is to change the winner by changing minimum number of votes. Dorn et al. [Dorn and Schlotter, 2012] studies the parameterized complexity of the Swap Bribery problem and Bredereck et al. [Bredereck et al., 2014] explores the parameterized complexity of the Shift Bribery problem for a wide range of parameters. We recall again that the costs and the budgets involved in all the bribery problems above need not necessarily correspond to actual money traded between voters and candidates. They may correspond to any cost in general, for example, the amount of effort or time that the briber needs to spend for each voter. ### 1.1 Motivation and Contributions In this work, we propose an unusual but effective cost model for the bribery problem. Even the most general cost models that have been studied in the literature fix absolute costs per voter-candidate combination, with no specific consideration to the voters’ opinions about the current winner. We introduce a model where a change of vote is relatively easier to effect if the change causes an outcome that the voter would find desirable. Indeed, if the currently winning candidate is, say, , and a voter is (truthfully) promised that by changing her vote from to , the winner of the election would change from to , then this is a change that the voter is likely to be happy to make. While the change does not make her favorite candidate win the election, it does improve the result from her point of view. Thus, given the circumstances (namely that of her least favorite candidate winning the election), the altered vote serves the voter better than the original one. We believe this perspective of voter influence is somewhat more sophisticated than the existing models in the literature. The cost of a change of vote is proportional to the nature of the outcome that the change promises — the cost is low or nil if the change results in better outcome with respect to the voter’s original ranking, and high or infinity otherwise. A frugal agent only approaches voters of the former category, thus being able to effectively bribe with minimal or no money. Formally, let be the winner of an election and let (other than ) the candidate whom the briber wishes to make the winner of the election. Now, the voters who prefer to will be reluctant to change their votes, and we call these votes non-vulnerable with respect to — we do not allow these votes to be changed by the briber, which justifies the frugal nature of the briber. On the other hand, if a voter prefers to , then it maybe very easy to convince her to change her vote if doing so makes win the election. We name these votes vulnerable with respect to . When is clear from the context, we simply call these votes non-vulnerable and vulnerable, respectively. The computational problem is to determine whether there are ways to make the candidate win the election by changing only those votes that are vulnerable with respect to . We call this problem Frugal-bribery. Note that there is no money involved in the Frugal-bribery problem. We also extend this basic model to a more general setting where each vulnerable vote has a certain nonnegative integer price which may correspond to the effort involved in approaching these voters and convincing them to change their votes. We also allow for the specification of a budget constraint, which can be used to model, for example, time constraints. This leads us to define the Frugal-$Bribery problem, where we are required to find vulnerable votes with a total cost that is within the budget; such that these votes can be changed in some way to make win the election. Note that the Frugal-$Bribery problem can be either uniform or nonuniform depending on whether the prices of the vulnerable votes are all identical or different. If not mentioned otherwise, the prices of the vulnerable votes will be assumed to be nonuniform. We remind that the briber is not allowed to change the non-vulnerable votes in both the Frugal-bribery and the Frugal-$Bribery problems.

We study and derive several interesting results on the computational complexity of the Frugal-bribery and the Frugal-$bribery problems with both uniform and nonuniform prices in unweighted as well as weighted elections for many commonly used voting rules. We believe that these problems are important as well as natural specializations of the$Bribery problem and their study has both theoretical and practical impact. The specific contributions of this paper are summarized below.

#### Unweighted Elections

We have the following results for unweighted elections.

• The Frugal-bribery problem is in for the -approval, Bucklin, and plurality with runoff voting rules. Also, the Frugal-$bribery problem is in for the plurality and veto voting rules. In contrast, the Frugal-$bribery problem is in -complete () for the Borda, maximin, Copeland, and STV voting rules [Lemma 3].

• The Frugal-bribery problem is in for the Borda voting rule [Theorem 1]. The Frugal-$bribery is in for the -approval for any constant [Theorem 2], -veto for any constant [Theorem 3], and a wide class of scoring rules [Theorem 5] even if the price of every vulnerable vote is either or . Moreover, the Uniform-frugal-$bribery is in for the Borda voting rule even if all the vulnerable votes have uniform price of and the budget is [Theorem 6].

• The Frugal-$bribery problem is in for the -approval, Bucklin, and plurality with runoff voting rules when the budget is a constant [Theorem 4]. #### Weighted Elections We have the following results for weighted elections. • The Frugal-bribery problem is in for the maximin and Copeland voting rules when we have only three candidates [Lemma 6], and for the plurality voting rule for any number of candidates [Theorem 7]. • The Frugal-bribery problem is in for the STV [Theorem 11], plurality with runoff [Corollary 1], and every scoring rule except the plurality voting rule [Lemma 7] for three candidates. The Frugal-$bribery problem is in for the plurality voting rule even for three candidates [Theorem 8].

• The Frugal-bribery problem is in for the maximin [Theorem 9], Bucklin [Theorem 12], and Copeland [Theorem 10] voting rules when we have only four candidates.

We summarize the results in the Table 1.

### 1.2 Related Work

(-Frugal-$bribery) Given a voting rule , a set of candidates , a preference profile , a distinguished candidate in , a finite budget , and a price function , determine if there is a way to make win the election by changing only the vulnerable votes subject to budget constraint . If the prices of all vulnerable votes are the same then we call the problem uniform-Frugal-$bribery. Otherwise, we call it nonuniform-Frugal-$bribery. The above problems are specializations of the well studied$Bribery problem. Also, the Coalitional-Manipulation problem [Bartholdi III et al., 1989, Conitzer et al., 2007], one of the classic problems in computational social choice theory, turns out to be a special case of the Frugal-$bribery problem [see Proposition 1]. For the sake of completeness, we include the definitions of these problems here. ###### Definition 4. (-$Bribery[Faliszewski et al., 2009a]
Given a voting rule , a set of candidates , a preference profile , a distinguished candidate in , a price function , and a budget , determine if there a way to make win the election.

###### Definition 5.

(Coalitional-Manipulation[Bartholdi III et al., 1989, Conitzer et al., 2007]
Given a voting rule , a set of candidates , a preference profile of truthful voters, an integer , and a distinguished candidate in , determine if there exists a voter preference profile such that the candidate wins uniquely (does not tie with any other candidate) in the profile .

We denote an arbitrary instance of the Coalitional-Manipulation problem by . The following proposition shows the relationship among the above problems.

###### Proposition 1.

For every voting rule, Frugal-bribery 222The notation denotes that the problem polynomial time many-to-one reduces to problem  [Garey and Johnson, 1979]. uniform-Frugal-$bribery nonuniform-Frugal-$bribery $Bribery. Also, Coalitional-Manipulation nonuniform-Frugal-$bribery.

###### Proof.

Given a Frugal-bribery instance, we construct a Uniform-frugal-$bribery instance by defining the price of every vulnerable vote to be zero and the budget to be zero. Clearly, the two instances are equivalent. Hence, Frugal-bribery Uniform-frugal-$bribery.

Uniform-frugal-$bribery Nonuniform-frugal-$bribery $Bribery follows from the fact that Uniform-frugal-$bribery is a specialization of Nonuniform-frugal-$bribery which in turn is a specialization of$Bribery.

Given a Manipulation instance, we construct a Nonuniform-frugal-$bribery instance as follows. Let be the distinguished candidate of the manipulators and be any arbitrary but fixed ordering of the candidates given in the Manipulation instance. Without loss of generality, we can assume that does not win if all the manipulators vote (Since, this is a polynomially checkable case of Manipulation). We define the vote of the manipulators to be , the distinguished candidate of the campaigner to be , the budget of the campaigner to be zero, the price of the manipulators to be zero and the price of the rest of the vulnerable votes to be one. Clearly, the two instances are equivalent. Hence, Manipulation Nonuniform-frugal-$bribery. ∎

###### Proof.

Given an algorithm for the Frugal-bribery problem, we iterate over all possible distinguished candidates to have an algorithm for the persuasion problem.

###### Proof.

The Manipulation problem is in for the above voting rules. Hence, the result follows from Proposition 1. ∎

###### Proof.

The problem is clearly in . To show -hardness, we reduce an arbitrary instance of X3C to Frugal-$bribery. Let be an instance of X3C. We define a Frugal-$bribery instance as follows. The candidate set is , where . For each , we add a vote as follows.

 vi:p≻q≻Si5 candidates≻D≻others

By Lemma 4, we can add many additional votes to ensure the following scores (denoted by ).

 s(p)=s(q)−|U|3,s(p)=s(x)−1,∀x∈U,
 s(p)>s(d)+|U|3,∀d∈D

The tie-breaking rule is “”. The winner is . The distinguished candidate is and thus all the votes in are vulnerable. The price of every is and the price of every other vulnerable vote is . The budget is . We claim that the two instances are equivalent. Suppose there exists an index set with such that . We replace the votes with as follows.

 v′i:p≻Dk candidates≻others

This makes the score of not less than the score of any other candidate and thus wins.

###### Proof.

The problem is clearly in . To show -hardness, we reduce an arbitrary instance of X3C to Frugal-$bribery. Let be any instance of X3C. We define a Frugal-$bribery instance as follows. The candidate set is , where . For each , we add a vote as follows.

 vi:p≻others≻Si≻Qk candidates

By Lemma 4, we can add many additional votes to ensure the following scores (denoted by ).

 s(p)>s(d),s(p)=s(x)+2,∀x∈U,s(p)=s(q)+1,∀q∈Q,
 s(p)=s(ai)−|U|3+1,∀i=1,2,3

The tie-breaking rule is “”. The winner is . The distinguished candidate is and thus all the votes in are vulnerable. The price of every is one and the price of any other vote is . The budget is . We claim that the two instances are equivalent. Suppose there exists an index set with such that . We replace the votes with as follows.

 v′i:others≻a1≻a2≻a3≻Qk candidates

The score of each decreases by and their final scores are , since the score of is not affected by this change. Also the final score of each is since forms an exact set cover. This makes win the election.

To prove the result in the other direction, suppose the Frugal-$bribery instance is a Yes instance. Then notice that there will be exactly votes in , where every , should come in the last positions since and the budget is . Notice that, candidates in must not be placed within top positions since , for every . Hence, in the votes that have been changed, and all the candidates in must occupy the last positions. We claim that the ’s corresponding to the ’s that have been changed must form an exact set cover. If not, then, there must exist a candidate and two votes and such that, both and have been replaced by and and the candidate was present within the last positions in both and . This makes the score of at least the score of which contradicts the fact that wins. ∎ However, the following result shows a polynomial time algorithm for the Frugal-$bribery problem for the -approval, Bucklin, and plurality with runoff voting rules, when the budget of the campaigner is a constant.

###### Theorem 5.

For any positional scoring rule with score vectors , if there exists a polynomial function such that, for every and in the score vector , there exists a satisfying the following condition:

 αi−αi+1=αi+1−αi+2>0,∀l≤i≤l+3

###### Proof.

The problem is clearly in . To show -hardness, we reduce an arbitrary instance of the Manipulation problem for the Borda voting rule with two manipulators to an instance of the uniform-Frugal-$bribery problem for the Borda voting rule. Let be an arbitrary instance of the Manipulation problem for the Borda voting rule. The corresponding Frugal-$bribery instance is as follows. The candidate set is . For each vote , we add a vote . Corresponding to the two manipulators’, we add two more votes , where is an arbitrary but fixed order of the candidates in . We add more votes to ensure the following score differences ( and are the score functions for the Manipulation and the uniform-Frugal-$bribery instances respectively).  s′(p)=λ+s(p)−2,s′(x)=λ+s(x) for every x∈C,  s′(q)=s′(p)−2m+1,s′(p)>s′(d)+2m for some% λ∈Z This will be achieved as follows. For any two arbitrary candidates and , the following two votes increase the score of by one more than the rest of the candidates except whose score increases by one less. This construction has been used before [Xia et al., 2010, Davies et al., 2011].  −→C1≻a≻b≻−−−−−−−−−−−−−→C∖(C1∪{a,b})  ←−−−−−−−−−−−−−C∖(C1∪{a,b})≻a≻b≻←−C1 Also we can ensure that candidate is always in positions and the candidate never immediately follows in these new votes. The tie-breaking rule is “”. The distinguished candidate is . The price of every vulnerable vote is one and the budget is two. We claim that the two instances are equivalent. Suppose the Manipulation instance is a Yes instance. Let be the manipulators’ votes that make win. In the Frugal-$bribery instance, we replace by for . This makes win the election. On the other hand, since in all the vulnerable votes except and , the candidate never immediately follows candidate , changing any of these votes can never make win the election since and the budget is two. Hence, the only way can win the election, if at all possible, is by changing the votes and . Let a vote replaces for . We can assume, without loss of generality, that and are at the first and the second positions respectively in both and . Let be the order restricted only to the candidates in . This makes winner of the Manipulation instance since, for every . ∎

## 4 Weighted Elections

Now, we turn our attention to weighted elections.

###### Lemma 5.

The following results follow immediately from the literature [Conitzer et al., 2007] and Proposition 1.

• The Frugal-bribery problem is in for the maximin and the Copeland voting rules for three candidates.

• The Frugal-bribery problem is in for any scoring rule except plurality for three candidates.

###### Lemma 6.

The Frugal-bribery problem is in for the maximin and the Copeland voting rules for three candidates.

###### Proof.

The Manipulation problem is in for the maximin, Copeland voting rules for three candidates [Conitzer et al., 2007]. Hence, the result follows from Proposition 1. ∎

###### Lemma 7.

The Frugal-bribery problem is in for any scoring rule except plurality for three candidates.

###### Proof.

The same proof for Theorem 6 of [Conitzer et al., 2007] would work here. ∎

###### Theorem 7.

The Frugal-bribery problem is in for the plurality voting rule.

###### Proof.

Let be the distinguished candidate of the campaigner. The greedy strategy of just replacing every vulnerable vote by solves the problem due to the monotonicity property of the plurality voting rule. ∎

The Partition problem, which is known to be [Garey and Johnson, 1979], is defined as follows.

###### Definition 7.

(Partition Problem)
Given a finite multi-set of positive integers with , does there exist a subset such that ? An arbitrary instance of Partition is denoted by .

We define another problem which we call -Partition as below and prove that it is also . We will use this fact in the proof of Theorem 11.

###### Definition 8.

(-Partition Problem)
Given a finite multi-set of positive integers with , does there exist a subset such that ? An arbitrary instance of -Partition is denoted by .

###### Lemma 8.

-Partition problem is in .

###### Proof.

The problem is clearly in . To show -hardness, we reduce the Partition problem to it. Let be an arbitrary instance of the Partition problem. We can assume, without loss of generality, that , since otherwise the instance is trivially a no instance. The corresponding -Partition problem instance is defined by , where . We claim that the two instances are equivalent. Suppose, the Partition instance is a Yes instance and thus there exists a set such that . This gives a solution to the -Partition instance. To prove the result in the other the other direction, suppose there is a set such that . This gives a solution to the Partition problem instance since . ∎

We now present the hardness results in weighted elections. We start with the plurality voting rule below.

The Frugal-$bribery problem is in for the plurality voting rule for three candidates. ###### Proof. The problem is clearly in . We reduce an arbitrary instance of Partition to an instance of Frugal-$bribery for the plurality voting rule. Let with and , be an arbitrary instance of the Partition problem. The candidates are and . We will add votes in such a way that makes win the election. The distinguished candidate is . For every , there is a vote of both weight and price . There are two votes of weight (we do not need to define the price of this vote since it is non-vulnerable) and of both weight and price . The tie-breaking rule is “”. We define the budget to be . We claim that the two instances are equivalent.

Suppose there exists a such that . We change the votes corresponding to the weights in to . This makes win the election with a score of . To prove in the other direction, for to win, its score must increase by at least . Also, the prices ensure that ’s score can increase by at most . Hence, ’s score must increase by exactly by and the only way to achieve this is to increase its score by changing the votes corresponding to the weights in . Thus, can win only if there exists a such that . ∎

Next we show the hardness result for the maximin voting rule.

###### Theorem 9.

The Frugal-bribery problem is in for the maximin voting rule for four candidates.

###### Proof.

The problem is clearly in . We reduce an arbitrary instance of Partition to an instance of Frugal-bribery for the maximin voting rule. Let with and , be an arbitrary instance of the Partition problem. The candidates are and . For every , there is a vote of weight . There is one vote , one , and one each of weight . The tie-breaking rule is “”. The distinguished candidate is . Let us call the set of votes corresponding to the weights in and the rest of the votes . Notice that, only the votes in are vulnerable. We claim that the two instances are equivalent.

Suppose there exists a such that . We change the votes corresponding to the weights in to . We change the rest of the votes in to . The maximin score of every candidate is and thus due to the tie-breaking rule, wins the election.

On the other hand, suppose there is a way to change the vulnerable votes, that is the votes in , that makes win the election. Without loss of generality, we can assume that all the votes in place at top position. First notice that the only way could win is that the vertices and must form a cycle in the weighted majority graph. Otherwise, one of and will be a Condorcet winner and thus the winner of the election. Now, we show that the candidate must defeat the candidate . If not, then must defeat my a margin of since the maximin score of is fixed at . Also, must defeat by a margin of , otherwise the maximin score of will be more than . This implies that all the votes in must be which makes defeat . This is a contradiction since the vertices and must form a cycle in the weighted majority graph. Hence must defeat with a margin of . This forces every vote in to prefer over . Hence, without loss of generality, we assume that all the votes in are either or , since whenever is right after , we can swap and and this will only reduce the score of without affecting the score of any other candidates. If the total weight of the votes in is more than , then , thereby making the maximin score of more than the maximin score of . If the total weight of the votes in is less than then, , thereby making the maximin score of more than the maximin score of . Thus the total weight of the votes in should be exactly which corresponds to a partition of . ∎

We turn our attention to the Copeland voting rule next.

###### Theorem 10.

The Frugal-bribery problem is in for the Copeland voting rule for four candidates, whenever .

###### Proof.

The problem is clearly in . We reduce an arbitrary instance of Partition to an instance of Frugal-bribery for the Copeland voting rule. Let with and , be an arbitrary instance of the Partition problem. The candidates are and . For every , there is a vote of weight . There are two votes and each of weight . The tie-breaking rule is “”. The distinguished candidate is . Let us call the set of votes corresponding to the weights in be and the rest of the votes be . Notice that, only the votes in are vulnerable. We claim that the two instances are equivalent.

Suppose there exists a such that . We change the votes corresponding to the weights in to . We change the rest of the votes in to . This makes win the election with a Copeland score of two.

On the other hand, suppose there is a way to change the votes in that makes win the election. Without loss of generality, we can assume that all the votes in place at top position. We will show that one of the three pairwise elections among and must be a tie. Suppose not, then must lose to both and , otherwise wins the election due to the tie-breaking rule. Now, consider the pairwise election between and . If defeats , then wins the election due to the tie-breaking rule. If defeats , then wins the election again due to the tie-breaking rule. Hence, one of the pairwise elections among and must be a tie. Without loss of generality suppose and ties. Then the total weight of the votes that prefer to in must be which constitutes a partition of . ∎

We now show the hardness result for the STV voting rule.

###### Theorem 11.

The Frugal-bribery problem is in for the STV voting rule for three candidates.

###### Proof.

The problem is clearly in . We reduce an arbitrary instance of -Partition to an instance of Frugal-bribery for the STV voting rule. Let with and , be an arbitrary instance of the -Partition problem. The candidates are and . For every , there is a vote of weight . There is a vote of weight and a vote of weight . The tie-breaking rule is “”. The distinguished candidate is . Let us call the set of votes corresponding to the weights in and the rest of the votes be . Notice that, only the votes in are vulnerable. We claim that the two instances are equivalent.

Suppose there exists a such that . We change the votes corresponding to the weights in to . We do not change the rest of the votes in . This makes win the election.

To prove the other direction, suppose there is a way to change the votes in that makes win the election. First notice that, can win only if qualifies for the second round. Hence, the total weight of the votes in that put at the first position must be at least . On the other hand, if the total weight of the votes in that put at the first position is strictly more than , then does not qualify for the second round and thus cannot win the election. Hence the total weight of the votes in that put at the first position must be exactly equal to which constitutes a -Partition of . ∎

For three candidates, the STV voting rule is the same as the plurality with runoff voting rule. Hence, we have the following corollary.

###### Corollary 1.

The Frugal-bribery problem is in for the plurality with runoff voting rule for three candidates.

Next we show, that the Frugal-bribery problem for the Bucklin voting rule is in .

###### Theorem 12.

The Frugal-bribery problem is in for the Bucklin voting rule for four candidates.

###### Proof.

The problem is clearly in . We reduce an arbitrary instance of Partition to an instance of Frugal-bribery for the Bucklin voting rule. Let with and , be an arbitrary instance of the Partition problem. The candidates are and . For every , there is a vote of weight . There are two votes and each of weight . The tie-breaking rule is “”. The distinguished candidate is . Let denotes the set of votes corresponding to the weights in and the rest of the votes be . Notice that, only the votes in are vulnerable. We claim that the two instances are equivalent.

Suppose there exists a such that . We change the votes corresponding to the weights in to . We change the rest of the votes in to . This makes win the election with a Bucklin score of three.

To prove the result in the other direction, suppose there is a way to change the votes in that makes win the election. Without loss of generality, we can assume that all the votes in place at the first position. First notice that the Bucklin score of is already fixed at three. In the votes in , the candidate can never be placed at the second position since that will make the Bucklin score of to be two. Also the total weight of the votes in that place in their second position can be at most . The same holds for . Hence, the total weight of the votes that place in their second position will be exactly equal to which constitutes a partition of . ∎

From Proposition 1, Lemma 7, Theorem 8 to 12 , and Corollary 1, we get the following corollary.

###### Corollary 2.

The uniform-Frugal-$bribery and the nonuniform-Frugal-$bribery problems are in for the scoring rules except plurality, STV, and the plurality with runoff voting rules for three candidates and for the maximin, Copeland, and Bucklin voting rules for four candidates.

## 5 Conclusion

We have proposed and studied a sophisticated model of bribery in this work. It turns out that the problems studied in this paper present interesting connections to two of the best studied problems in computational social choice theory, namely Manipulation and \$Bribery. We believe that our work has practical appeal as well while it is interesting, on its own, from a purely theoretical point of view.

A potential and natural direction for future work is to study these problems under various other settings. One obvious setting is to restrict the campaigner’s knowledge about the votes and/or the candidates who will actually turn up. The uncertainty can also arise from the voting rule that will eventually be used among a set of voting rules. Also, studying these problems when the pricing model for vulnerable votes is similar to swap bribery would be another interesting future direction. However, it is not very clear whether a swap bribery like pricing model has any practical motivation in the setting of bribery without money.

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