1 Introduction
In an election, there are a set of candidates and a set of voters. Each voter has a preference list of candidates. Given these preference lists, a winner is determined based on some voting rule, examples of which will be elaborated later.
In the context of election, the bribery problem has received considerable attention (see, for example, [1, 7, 8, 12, 14]). In this problem, there is an attacker who attempts to manipulate the election by bribing some voters, who will then report preference lists of the attacker’s choice (rather than the voters’ own preference lists). Each voter has a price for being bribed, and the attacker has an attack budget for bribing voters. There are two kinds of attackers: constructive vs. destructive. A constructive attacker attempts to make its designated candidate win an election, whereas the designated candidate would not win the election should there be no attacker. In contrast, a destructive attacker attempts to make its designated candidate lose the election, whereas the designated candidate would win the election should there be no attacker. The research question is: Given an attack budget for bribing, whether or not a (constructive or destructive) attacker can achieve its goal?
In this paper, we initiate the study of a new problem, called the protection problem, which extends the bribery problem as follows. There are also a set of candidates, a set of voters, and a bribery attacker. Each voter also has a preference list of candidates. There is also a voting rule according to which a winner is determined. Going beyond the bribery problem, the protection problem further considers a defender, who aims to protect elections from bribery. More specifically, the defender is given a defense budget and can use the defense budget to award some of the voters so that they cannot be bribed by the attacker anymore. This leads to an interesting problem: Given a defense budget, is it possible to protect an election from an attacker with a given attack budget for bribing voters (i.e., assuring that the attacker cannot achieve its goal)?
Our contributions. We introduce the problem of protecting elections from bribery, namely the protection problem. Given a defense budget for rewarding some of the voters and an attack budget for bribing some of the rest voters, the protection problem asks whether or not the defender can protect the election. We investigate the protection problem against the aforementioned two kinds of bribery attackers: constructive vs. destructive.
We present a characterization on the computational complexity of the protection problem (summarized in Table 1 in Section 5). The characterization is primarily concerning the voting rule of approval, which will be elaborated in Section 2. At a high level, our results can be summarized as follows. (i) The protection problem is hard and might be much harder than the bribery problem. For example, the protection problem is complete in most cases, while the bribery problem is in NP under the same settings. (ii) The destructive protection problem (i.e., protecting elections against a destructive attacker) is no harder than the constructive protection problem (i.e., protecting elections against a constructive attacker) in all of the settings we considered. In particular, the destructive protection problem is hard only when the voters are weighted and have arbitrary prices, while the constructive protection problem is hard even when the voters are unweighted and have the unit price. (iii) Voter weights and prices have completely different effects on the computational complexity of the protection problem. For example, the constructive protection problem is coNPhard in one case but is in P in another case.
Related Work. The problem of protecting elections from attacks seemingly has not received the due attention. Very recently, Yin et al. [20] considered the problem of defending elections against an attacker who can delete (groups of) voters. That is, the investigation is in the context of the control problem, where the attacker attempts to manipulate an election by adding or deleting some voters. The control problem has been extensively investigated (see, for example, [4, 9, 10, 20]). Although the control problem is related to the bribery problem, the means used by the attacker in the control problem (i.e., attacker adding or deleting some voters) is different from the means used by the attacker in the bribery problem (i.e., attacker changing the preference lists of the bribed voters). We investigate the protection problem, which is defined in the context of the bribery problem rather than the control problem. That is, the problem we investigate is different from the problem investigated by Yin et al. [20].
The protection problem we study is inspired by the bribery problem. Faliszewski et al. [8] gave the first characterization on the complexity of the bribery problem, including some dichotomy theorems. In the bribery problem, the attacker can pay a fixed, but voterdependent, price to arbitrarily manipulate the preference list of a bribed voter. The complexity of the bribery problem under the scoring rule of approval or veto for small values of was addressed later by Lin [15] and Bredereck and Talmon [2]. There are also studies on measuring the bribery price in different ways (see, e.g., [1, 6, 12]).
Technically, the protection problem is related to the bilevel optimization problem, especially the bilevel knapsack problem ([3, 5, 17, 18]). In the bilevel knapsack problem, there is a leader and a follower. The leader makes a decision first (e.g., packing a subset of items into the knapsack), and then the follower solves an optimization problem given the leader’s decision (e.g., finding the most profitable subset of items that have not been packed by the leader). The problem asks for the decision of the leader such that a certain objective function is optimized (e.g., minimizing the profit of the follower). The protection problem we study can be formulated as the bilevel problem by letting the defender award some voters who therefore cannot be bribed by the attacker anymore, and then the attacker bribes some of the remaining voters as an attempt to manipulate the election.
2 Problem definition
Election model. Consider a set of candidates and a set of voters . Each voter has a preference list of candidates, which is essentially a permutation of candidates, denoted as . The preference of is denoted by , meaning that prefers candidate to , where . Since is a permutation over , we denote by the inverse of , meaning that is the position of candidate
in vector
.Voting rules. In this paper, we focus on the scoring rule (or scoring protocol) that maps a preference list to a vector , where is the score assigned to the th candidate on the preference list of voter and . Given that is the preference list of , candidate receives a score of from . The total score of a candidate is the summation of the scores it received from the voters. The winner is the candidate that receives the highest total score. We focus on a singlewinner election, meaning that only one winner is selected. In the case of a tie, a random candidate with the highest total score is selected. However, our results remain valid for allnatural variation of selecting a single winner.
We say a scoring rule is nontrivial, if (i.e., not all scores are the same). There are many (nontrivial) scoring rules, including the popular approval, plurality, veto, Borda count and so on. In the case of approval, . In the case of plurality, . In the case of veto, . It is clear that plurality and veto are special cases of the scoring rule of approval.
Weights of voters. Voters can have different weights. Let be the weight of voter . In a weighted election, the total score of a candidate is the weighted sum of the scores a candidate receives from the voters. For example, candidate receives a score from voter .
By reindexing all of the candidates, we can set, without loss of generality, as the winner in the absence of bribery.
Adversarial models. We consider an attacker that does not belong to but attempts to manipulate the election by bribing some voters. Suppose voter has a bribing price , meaning that , upon receiving a bribery of amount from the attacker, will change its preference list to the list given by the attacker. The attacker has a total budget . As in the bribery problem, we also consider two kinds of attackers:

Constructive attacker: This attacker attempts to make a designated candidate win the election, meaning that the designated candidate is the only candidate who gets the highest score.

Destructive attacker: This attacker attempts to make a designated candidate lose the election, meaning that there is another candidate that gets a strictly higher score than the designated candidate does.
Protection. In the protection problem, voter , upon receiving an award of amount (or awarding price) from the defender, will always report its preference list faithfully and cannot be bribed. Note that may have multiple interpretations, such as monetary award, economic incentives or the cost of isolating voters from bribery. We say a voter is awarded if receives an award of .
Problem statement. We formalize our problem as follows.
The constructive protection problem (i.e., protecting elections against constructive attackers): Input: A set of candidates. A set of voters, each with a weight , a preference list , an awarding price of and a bribing price of . A scoring rule for selecting a single winner. A defender with a defense budget . An attacker with an attack budget attempting to make candidate win the election. Output: Decide whether there exists a such that ; and for any subset with , does not get a strictly higher score than any other candidate despite the attacker bribing the voters belonging to (i.e., bribing ).
The destructive protection problem (i.e., protecting elections against destructive attackers): Input: A set of candidates. A set of voters, each with a weight , a preference list , an awarding price of and a bribing price of . A scoring rule for selecting a single winner. Suppose is the winner if no voter is bribed. A defender with a defense budget . An attacker with an attack budget attempting to make lose the election by making get a strictly higher score than does. Output: Decide if there exists a such that ; and for any subset such that , no candidate can get a strictly higher score than does despite the attacker bribing .
Further terminology and notations. We denote by the total score obtained by candidate in the absence of bribery (i.e., no voter is bribed). If the defender can select such that no constructive or destructive attacker can succeed, we say the defender succeeds. We call our problem as the (constructive or destructive) weighted$protection problem, where “weighted” indicates that the voters are weighted and “$” indicates that arbitrary awarding and bribing prices are involved. In addition to investigating the general weighted$protection problem, we also investigate the following special cases of it:

the $protection problem with for each (i.e., the voters are not weighted);

the weightedprotection problem with for each (i.e., voters are associated with the unit awarding price and the unit bribing price);

the unitprotection problem with for each (i.e., voters are not weighted, and are associated with the unit awarding price and the unit bribing price).

the symmetric protection problem with for each (i.e., the awarding price and the bribing price are always the same), while noting that different voters may have different prices.
3 The Case of Constantly Many Candidates
3.1 The Weighted$Protection Problem
The goal of this subsection is to prove the following theorem.
Theorem 3.1
For any nontrivial scoring rule, both the constructive and destructive weighted$protection problem, is complete.
The theorem follows from Lemma 1 and Lemma 2 below, which shows the membership and hardness, respectively.
Lemma 1
For any nontrivial scoring rule, both the constructive and destructive weighted$protection problems are in .
Lemma 2
For any nontrivial scoring rule, both the constructive and destructive weighted$protection problems are both hard even if there are only candidates.
Proof sketch. The proof of Lemma 2 follows from DeNegre (DNeg) variant of bilevel knapsack problem, which is proved to be hard by Caprara et al. [3]. We give a brief explanation. In this DeNegre variant, there are an adversary and a packer. The adversary has a reserving budget and the packer has a packing budget . There is a set of items, each having a price to the adversary, a price to the packer, and a weight to both the adversary and the packer. The adversary first reserves a subset of items whose total price is no more than . Then the packer solves the knapsack problem with respect to the remaining items that are not reserved; i.e., the packer will select a subset of the remaining items whose total price is no more than but their total weight is maximized. The DeNegre variant asks if the adversary can reserve a proper subset of items such that the total weight of the unreserved items that are selected by the packer is no more than some parameter . The DeNegre variant is similar to the weighted$protection protection problem, because we can view the defender and attacker in the protection problem respectively as the adversary and packer in the bilevel knapsack problem. In the case of a singlewinner election with candidates, the goal of the defender is to assure that the constructive attacker cannot make the loser get a strictly higher score than the winner by bribing. This is essentially the same as ensuring that the constructive attacker cannot bribe a subset of nonawarded voters whose total weight is higher than a certain threshold, which is the same as the bilevel knapsack problem. ∎
3.2 The WeightedProtection Problem
This is a special case of the weighted$protection problem when .
The following theorem used by Faliszewski et al. [8] was originally proved for another problem. In our context, and thus , it is NPhard to decide if the constructive attacker can succeed or, equivalently, if the defender cannot succeed. Hence, it is coNPhard to decide if the defender can succeed and Theorem 3.2 follows.
Theorem 3.2
(By Faliszewski et al. [8]) If is a constant, the constructive weightedprotection problem is coNPhard for any scoring rule that are not all equal (i.e., it does not hold that ).
In contrast, the destructive version is easy. Using the fact that , the number of candidates, is a constant, we can prove the following Theorem 3.3 through suitable enumerations.
Theorem 3.3
If is a constant, then the destructive weightedprotection problem is in P for any scoring rule.
3.3 The $Protection Problem
This is the special case of the protection problem with for every . The following two theorems illustrate the significant difference (in terms of complexity) between the general problem and its special case with symmetricity (i.e., ).
Theorem 3.4
For constant and any nontrivial scoring rule, both the constructive and destructive $protection problems are NPcomplete.
Theorem 3.5
For constant , both destructive and constructive symmetric $protection problems are in for any scoring rule.
4 The Case of Arbitrarily Many Candidates
4.1 The Case of Constructive Attacker
The following Theorem 4.1 shows hardness for the most special cases of the constructive weighted$protection problem, namely (unitprotection). It thus implies readily the hardness for the more general constructive $protection and constructive weightedprotection.
Theorem 4.1
For arbitrary , the approval constructive unitprotection problem is complete.
Membership in follows directly from Lemma 1. To prove Theorem 4.1, it suffices to show the following.
Lemma 3
For arbitrary , the approval constructive unitprotection problem is hard even if .
To prove Lemma 3, we reduce from a variant of the 3 dimensional matching problem (or 3DM), which is called 3DM and defined below. The classical 3DM is hard proved by Mcloughlin [16]. By leveraging the proof by Mcloughlin [16], we can show the hardness of the 3DM problem.
3DM: Given a parameter , three disjoint sets of elements , , of the same cardinality, and two disjoint subsets and of such that contains each element of at most once. Does there exist a subset such that and for any , is not a perfect matching (where a perfect matching is a subset of triples in which every element of appears exactly once)?
Proof (Proof of Lemma 3)
Given an arbitrary instance of 3DM, we construct an instance of the constructive unitprotection problem in approval election as follows. Recall that and thus every voter votes for 4 candidates.
Suppose , , .
There are key candidates, including:

key candidates, each corresponding to one distinct element of and we call them element candidates. The score of every element candidate is ;

one key candidate called leading candidate, whose total score is ;

one key candidate called designated candidate, whose total score is .
Here is some sufficiently large integer, e.g., we can choose . Besides key candidates, there are also many dummy candidates, each of score either or . The number of dummy candidates will be determined later.
There are key voters, including:

key voters, each corresponding to a distinct triple in and we call them voters. For each , the corresponding voter votes for the candidates corresponding to elements , , together with the leading candidate;

key voters, each distinct triple in corresponds to exactly voters and we call them voters. For every , each of its corresponding voters vote for the candidates corresponding to elements , , together with one distinct dummy candidate. Since the voters are identical, we can view them as copies, i.e., every voter has copies.
Besides key voters, there are also sufficiently many dummy voters. Each dummy voter votes for exactly one key candidate and 3 distinct dummy candidates. Dummy voters and dummy candidates are used to make sure that the score of key candidates are exactly as we have described. More precisely, if we only count the scores of key candidates contributed by key voters, then the element candidate corresponding to has a score of where is the number of occurrences of in the triple set for , and the leading candidate has a score of . Hence, there are exactly dummy voters who vote for the element candidate corresponding to , and dummy voters who vote for the leading candidate.
Overall, we create dummy voters, and dummy candidates.
As the leading candidate is the current winner, the constructive unitprotection problem asks whether the election can be protected against an attacker attempting to make the designated candidate win. The defense budget is and the attack budget is . In the following we show that the defender succeeds if and only if the given 3DM instance admits a feasible solution .
“Yes” Instance of 3DM “Yes” Instance of Constructive UnitProtection. Suppose the instance of 3DM admits a feasible solution , we show that the answer for constructive unitprotection problem is “Yes”.
Recall that each voter corresponds to a distinct triple in and votes for 4 candidates – the leading candidate and the three candidates corresponding to . We do not award voters corresponding to the triples in , but award all of the remaining voters. The resulting cost is exactly . In what follows we show that after awarding voters this way, the attacker cannot make the designated candidate win.
Suppose on the contrary, the attacker can make the designated candidate win by bribing voters among the voters, voters among the voters, and dummy voters. We claim that the following inequalities hold:
(1a)  
(1b) 
Inequality (1a) follows from the fact that the attack budget is and the attacker can bribe at most voters. Inequality (1b) holds because of the following. Given that a candidate can get at most one score from each voter and that the attacker can bribe at most voters, bribing voters can make the designated candidate obtain a score at most . Hence, the score of each key candidate other than the designated one should be at most . Recall that without bribery, each of the element candidate has a score of and the leading candidate has a score of . Hence, the attacker should decrease at least 1 score from each element candidate and scores from the leading candidate, leading to a total score of . Note that an voter contributes 1 score to 4 key candidates, therefore it contributes in total a score of to the key candidates. Similarly an voter contributes a score of , and a dummy voter contributes a score of to the key candidates. Therefore, by bribing (for example) an voter, the total score of all the element candidates and the leading candidate can decrease by at most . Thus, inequality (1b) holds.
In the following we derive a contradiction based on Inequalities (1a) and (1b). By plugging into Inequality (1b), we have Since , we have . Hence, , and we have and . Note that the defender has awarded every voter except the ones corresponding to , where . Hence, every voter corresponding to the triples in is bribed. Furthermore, as Inequality (1b) is tight, bribing voters makes the designated candidate have a score of , while making each of the other key candidates have a score of . This means that the score of each element candidate decreases exactly by . Hence, the attacker has selected a subset of voters such that together with the voters corresponding to triples in , these voters contribute exactly a score of 1 to every element candidate. Let be the set of triples to which the bribed voters correspond, then forms a 3dimensional matching, which is a contradiction to the fact that is a feasible solution to the 3DM instance. Thus, the attacker cannot make the designated candidate win and the answer for the constructive unitprotection problem is “Yes”.
No Instance of 3DM No Instance of Constructive UnitProtection. Suppose for any , there exists such that is a perfect matching, we show that the answer to the constructive unitprotection problem is “No”. Consider an arbitrary set of voters awarded by the defender. Among the awarded voters, let be the set of triples that corresponds to the awarded voters. As , . We select an arbitrary subset such that . There exists some such that is a perfect matching, and we let the attacker bribe the set of voters corresponding to triples in . Note that this is always possible as every voter has copies, so no matter which voters are awarded the briber can always select one voter corresponding to each triple in . It is easy to see that by bribing these voters, the score of every element candidate decreases by , and the score of the leading voter decreases by . Meanwhile, let each bribed voter vote for the designated candidate and three distinct dummy candidates, then the designated candidate has a score of and becomes a winner, i.e., the answer to the constructive unitprotection problem is “No”. ∎
Remark. The proof of Lemma 3 can be easily modified to prove the hardness of approval constructive unitprotection problem for any fixed . Specifically, we can make the same reduction, and add dummy candidates such that every voter additionally votes for exactly distinct dummy candidates.
4.2 The Case of Destructive Attacker
Theorem 4.2
Both approval destructive weightedprotection and approval (symmetric) $protection problems are NPcomplete.
5 Summary of Results
The preceding characterization of the computational complexity of the protection problem in various settings is summarized in Table 1.
# of candidates  Model parameters  Destructive attacker  Constructive attacker 
constant  Weighted, Priced, Asymmetric  complete (Thm 3.1)  complete (Thm 3.1) 
Weighted,  P (Thm 3.3)  coNPhard (Thm 3.2)  
, Priced, Asymmetric  NPcomplete (Thm 3.4)  NPcomplete (Thm 3.4)  
, Priced, Symmetric  P (Thm 3.5)  P (Thm 3.5)  
,  P (Thm 3.5)  P (Thm 3.5)  
arbitrary  Weighted, Priced, Asymmetric  complete (Thm 3.1)  complete (Thm 3.1) 
Weighted,  NPcomplete (Thm 4.2)  hard (Thm 4.1)  
, Priced, Asymmetric  NPcomplete (Thm 4.2)  hard (Thm 4.1)  
, Priced, Symmetric  NPcomplete (Thm 4.2)  hard (Thm 4.1)  
,  ?  hard (Thm 4.1) 
We remark three natural open problems for future research. One is the complexity of the destructive protection problem with . It is not clear whether the problem is in P or is NPcomplete. Another is the constructive protection problem with and arbitrary voter weights. We only show its coNPhardness, it is not clear whether or not this problem is coNPcomplete. The third problem is the complexity of approval constructive unitprotection problem when as our hardness proof only holds when .
6 Conclusion
We introduced the protection problem and characterized its computational complexity. We showed that the problem, in general, is complete, and identified settings in which the problem becomes easier. Moreover, we showed the protection problem in some parameter settings is polynomialtime solvable, suggesting that these parameter settings can be used for realwork election applications.
In addition to the open problems mentioned in Section 5, the following are also worth investigating. First, our hardness results would motivate the study of approximation or FPT (fixed parameter tractable) algorithms for the protection problem. Note that even polynomial time approximation schemes can exist for hard problems (see, e.g., By Caparara et al. [3]). It is thus desirable that a similar result can be obtained for some variants of the protection problem. Second, how effective is this approach when applied towards the problem of defending against other types of attackers that can, e.g., add or delete votes? Third, much research remains to be done in extending the protection problem to accommodate other scoring rules such as Borda and Copeland.
7 Appendix
7.1 Destructive WeightedProtection – an Equivalent Formulation
We provide an equivalent formulation of the destructive weightedprotection problem under any scoring rule , which will be very useful for several proofs throughout this paper.
The minmax vector addition problem: Input: A vector where is the score of in the absence of bribery. An vector for each voter where . Awarding price and bribing price for voter , . Defense budget and attack budget . Output: Decide if there exists a subset such that ; and For any subset with , it holds that
Lemma 4
The answer to the destructive weighted$protection problem is “Yes” if and only if the answer to the corresponding minmax vector addition problem is “Yes”.
Proof
A “Yes” Instance of Minmax Vector Addition A “Yes” Instance of Destructive Weighted$Protection. Suppose the answer to the minmax vector addition problem is “Yes.” Then, there exists some such that for any with , it holds that
(2) 
For showing a contradiction, suppose the answer for the destructive weighted$protection problem is “No”. In this case, even if the defender awards the voters in , the attacker can still make lose by bribing some subset of voters. Note that if does not win, there must exist some other candidate, say, , who gets a strictly higher score than after the attacker bribes some voters. Let us compare their scores before and after bribing voters. Before bribing voters, the scores of and are and , respectively. Recall that a candidate is at the position of on the preference list of , therefore any contributes a score of to , and contributes a score of to . After bribing voters, the preference list of is changed, but regardless of the change, contributes at least to and at most to . Let the scores of and after bribing voters be and , respectively. Then, it follows that
Since , we have
that is,
which contradicts Eq. (2). Thus, the answer to the destructive weighted$protection problem is “Yes”.
A “Yes” Instance of Destructive Weighted$Protection A “Yes” Instance of Minmax Vector Addition. Suppose the answer to the destructive weighted$protection problem is “Yes” by awarding the voters in . We show that the answer to the corresponding instance of minmax vector addition problem is “Yes”. Suppose on the contrary the answer is “No.” Then, for there exists some such that
Consequently, there must exist some such that . By plugging in , we have
This means that if the defender awards the voters in , then the attacker can bribe the voters in to change their preference lists such that for any , candidate is on top of the list and is at bottom of the list. By doing this, gets a strictly higher score than . This contradicts the fact that the answer to the destructive weighted$protection problem is “Yes”. Hence, the answer to the minmax vector addition problem is “Yes”. ∎
7.2 Voter Dominance and Preliminary Observations
Let be the set of permutations over . Each element of can be a preference list. Let be the set of voters whose preference list is the th element of .
For two voters and , we say dominates (or is dominated by ), denoted by , if any of the following two conditions hold: (i) The following holds and at least one of the inequalities is strict:
(ii) The following holds:
Note that the domination relation is only defined between the voters who have the same preference. Intuitively, if , then is more “important” than because has a greater weight but is “cheaper” to bribe or award (i.e., more valuable to both the attacker and the defender).
We have the following lemmas.
Lemma 5
Consider the destructive weighted$protection problem with being the set of awarded voters. Suppose the attacker can succeed by bribing a subset of voters. If , and , then the attacker can succeed by bribing .
Towards the proof, we need Lemma 4, which states that the destructive weighted$protection problem is equivalent to another problem called minmax vector addition. The reader may refer to Section 4.2 for the definition of this problem. The following observation follows directly from the definition of which is included in the definition of minmax vector addition.
Observation 1
If , then for .
Proof (Proof of Lemma 5)
We prove the lemma by applying an exchange argument to the minmax vector addition problem, which is equivalent to the destructive weighted$protection by Lemma 4. Suppose by bribing voters in the destructive attacker can make lose. Then, it follows that and
As dominates , and . Hence,
and
That is, the briber can also win by bribing voters in . ∎
Lemma 6
Consider the destructive weighted$protection problem. Suppose the defender succeeds by awarding a subset of voters. If , and , then the defender can succeed by awarding .
Proof
We again use an exchange argument to the minmax vector addition problem. Let . Suppose on the contrary that the defender cannot win by fixing voters in , then there exists some such that and
We argue that the defender cannot win either by fixing voters in , which is a contradiction. Suppose the defender fixes voters in . There are two possibilities. If , then we let the briber bribe voters in . It is obvious that the briber can win. Otherwise, , then we let the briber bribe voters in . Since dominates , we have and
Hence, the lemma is true. ∎
We say is maximal (with respect to ) if for any , there is no that can dominate . That is, contains the most important voters. The following corollary follows directly from the preceding two lemmas.
Corollary 1
Consider the destructive weighted$protection problem. Without loss of generality, we can assume that is maximal with respect to and is maximal with respect to .
Unfortunately, Corollary 1 does not hold for the constructive weighted$protection problem, which is significantly different from the destructive version of the protection problem in terms of computational complexity. Nevertheless, similar results hold for the unweighted constructive problem.
Lemma 7
Given as the set of fixed voters in the constructive $briberyprotection problem, suppose a briber can make win by bribing a subset of voters. If , and , then the briber can also win by bribing voters in .
Proof
Again, we prove by an exchange argument. Suppose by bribing voters in the constructive briber can make win. Now we consider the following procedure: we change the preference list of into the same one as that of , and meanwhile, restore the preference list of to the original one. As voters have the same weight, this procedure does not change the total score of every candidate, and is thus still the winner. Furthermore, this procedure is equivalent as we bribe . Since dominates , the total cost of bribing is no more than that of bribing . Hence, the lemma is true. ∎
Lemma 8
In the constructive $briberyprotection problem, suppose the defender can win by fixing a subset of voters. If , and , then the defender can also win by fixing voters in .
Proof
Suppose on the contrary that the defender cannot win by fixing voters in . Then the constructive briber can win by bribing voters in some subset . There are two possibilities. If , then even if the defender fixes the briber can still bribe and make win, which is a contradiction. Otherwise . If the defender fixes , we let the briber bribe . According to Lemma 7, if the briber can win by bribing , he/she can also win by bribing , again contradicting the fact that the defender can succeed by awarding . ∎
The above Lemmas implies the following.
Corollary 2
Without loss of generality, we can assume that is maximal with respect to , and is maximal with respect to in the constructive $briberyprotection problem.
7.3 Proofs Omitted in Section 3.1
Recall that our goal is to prove Theorem 3.1.
Theorem 3.1
For any nontrivial scoring rule, both the constructive and destructive weighted$protection problem, is complete.
We first show membership.
Lemma 1
For any nontrivial scoring rule, both the constructive and destructive weighted$protection problems are in .
For ease of proof, we use the following definition of from [19] (see Theorem 3 therein).
Definition 1
(By Wrathall [19]) Let be a finite set of symbols (alphabet) and be the set of strings of symbols in . Let be a language. if and only if there exists polynomials , and a language such that for all ,
where denotes
Proof (Proof of Lemma 1)
Given an instance of the constructive or destructive weighted$protection problem, we want to know if there exists a subset such that and for any subset with and any preference list for , the following property is true: By bribing voter in and change the preference list of each to , a constructive attacker cannot make candidate win, or a destructive attacker cannot make lose. It is easy to see that the property
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