Computational Complexity Characterization of Protecting Elections from Bribery
Lin Chen, Ahmed Sunny, Lei Xu, Shouhuai Xu, Zhimin Gao, Yang Lu, Weidong Shi, Nolan Shah
aa r X i v : . [ c s . CC ] J u l Computational Complexity Characterization ofProtecting Elections from Bribery ∗ Lin Chen , Ahmed Sunny Lei Xu , Shouhuai Xu , Zhimin Gao , Yang Lu ,Weidong Shi , and Nolan Shah Texas Tech University, 2500 Broadway, Lubbock, TX 79409, USA [email protected] [email protected] University of Texas Rio Grande Valley,1201 W University Dr, Edinburg, TX 78539,USA [email protected] University of Texas San Antonio,1 UTSA Circle, San Antonio, TX 78249, USA [email protected] Auburn University at Montgomery,7430 East Dr, Montgomery, AL 36117, USA [email protected] University of Houston, 4800 Calhoun Rd, Houston, TX 77004, USA [email protected] [email protected] Amazon Web Services, Seatle, USA [email protected]
Abstract.
The bribery problem in election has received considerableattention in the literature, upon which various algorithmic and complex-ity results have been obtained. It is thus natural to ask whether we canprotect an election from potential bribery. We assume that the protectorcan protect a voter with some cost (e.g., by isolating the voter from po-tential bribers). A protected voter cannot be bribed. Under this setting,we consider the following bi-level decision problem: Is it possible for theprotector to protect a proper subset of voters such that no briber witha fixed budget on bribery can alter the election result? The goal of thispaper is to give a full picture on the complexity of protection problems.We give an extensive study on the protection problem and provide algo-rithmic and complexity results. Comparing our results with that on thebribery problems, we observe that the protection problem is in generalsignificantly harder. Indeed, it becomes Σ p -complete even for very re-stricted special cases, while most bribery problems lie in NP. However,it is not necessarily the case that the protection problem is always harder.Some of the protection problems can still be solved in polynomial time,while some of them remain as hard as the bribery problem under thesame setting. Keywords:
Voting, complexity, NP-hardness, Σ p -hardness ∗ A 2 page extended abstract has been published at the Proceedings of the17th International Conference on Autonomous Agents and MultiAgent Systems(AAMAS’18)
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 determinedbased on some voting rule, examples of which will be elaborated later.In the context of election, the bribery problem has received considerable at-tention (see, for example, [1,7,8,12,14]). In this problem, there is an attacker whoattempts to manipulate the election by bribing some voters, who will then reportpreference lists of the attacker’s choice (rather than the voters’ own preferencelists). Each voter has a price for being bribed, and the attacker has an attackbudget for bribing voters. There are two kinds of attackers: constructive vs. de-structive . A constructive attacker attempts to make its designated candidate winan election, whereas the designated candidate would not win the election shouldthere be no attacker. In contrast, a destructive attacker attempts to make itsdesignated candidate lose the election, whereas the designated candidate wouldwin the election should there be no attacker. The research question is: Given anattack budget for bribing, whether or not a (constructive or destructive) attackercan achieve its goal?In this paper, we initiate the study of a new problem, called the protectionproblem , which extends the bribery problem as follows. There are also a setof 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 awinner is determined. Going beyond the bribery problem, the protection problemfurther considers a defender, who aims to protect elections from bribery. Morespecifically, the defender is given a defense budget and can use the defense budgetto award some of the voters so that they cannot be bribed by the attackeranymore. This leads to an interesting problem:
Given a defense budget, is itpossible to protect an election from an attacker with a given attack budget forbribing voters (i.e., assuring that the attacker cannot achieve its goal)?
Our contributions.
We introduce the problem of protecting elections frombribery, namely the protection problem. Given a defense budget for rewardingsome of the voters and an attack budget for bribing some of the rest voters, theprotection problem asks whether or not the defender can protect the election.We investigate the protection problem against the aforementioned two kinds ofbribery attackers: constructive vs. destructive.We present a characterization on the computational complexity of the pro-tection problem (summarized in Table 1 in Section 5). The characterization isprimarily concerning the voting rule of r -approval, which will be elaborated inSection 2. At a high level, our results can be summarized as follows. (i) The pro-tection problem is hard and might be much harder than the bribery problem . Forexample, the protection problem is Σ p -complete in most cases, while the briberyproblem is in NP under the same settings. (ii) The destructive protection prob-lem (i.e., protecting elections against a destructive attacker) is no harder thanthe constructive protection problem (i.e., protecting elections against a construc-tive attacker) in all of the settings we considered. In particular, the destructiverotection problem is Σ p -hard only when the voters are weighted and have ar-bitrary prices, while the constructive protection problem is Σ p -hard even whenthe voters are unweighted and have the unit price. (iii) Voter weights and priceshave completely different effects on the computational complexity of the protec-tion problem. For example, the constructive protection problem is coNP-hard inone case but is in P in another case. Related Work.
The problem of protecting elections from attacks seeminglyhas not received the due attention. Very recently, Yin et al. [20] considered theproblem of defending elections against an attacker who can delete (groups of)voters. That is, the investigation is in the context of the control problem , wherethe attacker attempts to manipulate an election by adding or deleting somevoters. 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 addingor deleting some voters) is different from the means used by the attacker inthe bribery problem (i.e., attacker changing the preference lists of the bribedvoters). We investigate the protection problem, which is defined in the contextof the bribery problem rather than the control problem. That is, the problemwe investigate is different from the problem investigated by Yin et al. [20].The protection problem we study is inspired by the bribery problem. Fal-iszewski et al. [8] gave the first characterization on the complexity of the briberyproblem , including some dichotomy theorems. In the bribery problem, the at-tacker can pay a fixed, but voter-dependent, price to arbitrarily manipulate thepreference list of a bribed voter. The complexity of the bribery problem underthe scoring rule of r -approval or r -veto for small values of r was addressed laterby Lin [15] and Bredereck and Talmon [2]. There are also studies on measuringthe bribery price in different ways (see, e.g., [1,6,12]).Technically, the protection problem is related to the bi-level optimization problem, especially the bi-level knapsack problem ([3,5,17,18]). In the bi-levelknapsack problem, there is a leader and a follower. The leader makes a decisionfirst (e.g., packing a subset of items into the knapsack), and then the followersolves an optimization problem given the leader’s decision (e.g., finding the mostprofitable subset of items that have not been packed by the leader). The problemasks for the decision of the leader such that a certain objective function is op-timized (e.g., minimizing the profit of the follower). The protection problem westudy can be formulated as the bi-level problem by letting the defender awardsome voters who therefore cannot be bribed by the attacker anymore, and thenthe attacker bribes some of the remaining voters as an attempt to manipulatethe election. Election model.
Consider a set of m candidates C = { c , c , . . . , c m } and aset of n voters V = { v , v , . . . , v n } . Each voter v j has a preference list of can-didates, which is essentially a permutation of candidates, denoted as τ j . Thereference of v j is denoted by ( c τ j (1) , c τ j (2) , . . . , c τ j ( m ) ), meaning that v j preferscandidate c τ j ( z ) to c τ j ( z +1) , where z = 1 , , . . . . Since τ j is a permutation over { , , . . . , m } , we denote by τ − j the inverse of τ j , meaning that τ − j ( i ) is theposition of candidate c i in vector ( c τ j (1) , c τ j (2) , . . . , c τ j ( m ) ). Voting rules.
In this paper, we focus on the scoring rule (or scoring protocol)that maps a preference list to a m -vector α = ( α , α , . . . , α m ), where α i ∈ N is the score assigned to the i -th candidate on the preference list of voter v j and α ≥ α ≥ . . . ≥ α m . Given that τ j is the preference list of v j , candidate c τ j ( i ) receives a score of α i from v j . The total score of a candidate is the summation ofthe scores it received from the voters. The winner is the candidate that receivesthe highest total score. We focus on a single-winner election, meaning that onlyone winner is selected. In the case of a tie, a random candidate with the highesttotal score is selected. However, our results remain valid for all-natural variationof selecting a single winner.We say a scoring rule is non-trivial , if α > α m (i.e., not all scores arethe same). There are many (non-trivial) scoring rules, including the popular r -approval , plurality , veto , Borda count and so on. In the case of r -approval, α = (1 , , . . . , | {z } r , , , . . . , | {z } m − r ). In the case of plurality, α = (1 , , . . . , α = (1 , , . . . , , r -approval. Weights of voters.
Voters can have different weights. Let w j ∈ N be the weightof voter v j . In a weighted election, the total score of a candidate is the weightedsum of the scores a candidate receives from the voters. For example, candidate c i receives a score w j · α τ − j ( i ) from voter v j .By re-indexing all of the candidates, we can set, without loss of generality, c m as the winner in the absence of bribery. Adversarial models.
We consider an attacker that does not belong to
C ∪ V but attempts to manipulate the election by bribing some voters. Suppose voter v j has a bribing price p bj , meaning that v j , upon receiving a bribery of amount p bj from the attacker, will change its preference list to the list given by the attacker.The attacker has a total budget B . As in the bribery problem, we also considertwo kinds of attackers: – Constructive attacker : This attacker attempts to make a designated candi-date win the election, meaning that the designated candidate is the onlycandidate who gets the highest score. – Destructive attacker : This attacker attempts to make a designated candidatelose the election, meaning that there is another candidate that gets a strictlyhigher score than the designated candidate does.
Protection.
In the protection problem, voter v j , upon receiving an award ofamount p aj (or awarding price ) from the defender, will always report its pref-erence list faithfully and cannot be bribed. Note that p aj may have multipleinterpretations, such as monetary award, economic incentives or the cost of iso-ating voters from bribery. We say a voter v j is awarded if v j receives an awardof p aj . Problem statement.
We formalize our problem as follows.
The constructive protection problem (i.e., protecting electionsagainst constructive attackers):
Input:
A set C of m candidates. A set V of n voters, each with a weight w j ∈ Z > , a preference list τ j , an awarding price of p aj ∈ Z > and a bribing price of p bj ∈ Z > . A scoring rule for selecting a single winner. A defender with a defensebudget F ∈ Z ≥ . An attacker with an attack budget B ∈ Z ≥ attempting tomake candidate c m win the election. Output:
Decide whether there exists a V F ⊆ V such that – P j : v j ∈V F p aj ≤ F ; and – for any subset V B ⊆ V \ V F with P j : v j ∈V B p bj ≤ B , c m does not get astrictly higher score than any other candidate despite the attacker bribingthe voters belonging to V B (i.e., bribing V B ). The destructive protection problem (i.e., protecting elections againstdestructive attackers):
Input:
A set C of m candidates. A set V of n voters, each with a weight w j ∈ Z > , a preference list τ j , an awarding price of p aj ∈ Z > and a bribing price of p bj ∈ Z > . A scoring rule for selecting a single winner. Suppose c m is the winnerif no voter is bribed. A defender with a defense budget F ∈ Z ≥ . An attackerwith an attack budget B ∈ Z ≥ attempting to make c m lose the election bymaking c ∈ C \ { c m } get a strictly higher score than c m does. Output:
Decide if there exists a V F ⊆ V such that – P j : v j ∈V F p aj ≤ F ; and – for any subset V B ⊆ V \ V F such that P j : v j ∈V B p bj ≤ B , no candidate c ∈ C \ { c m } can get a strictly higher score than c m does despite theattacker bribing V B . Further terminology and notations.
We denote by W ( c i ) the total scoreobtained by candidate c i in the absence of bribery (i.e., no voter is bribed). Ifthe defender can select V F such that no constructive or destructive attacker cansucceed, we say the defender succeeds. We call our problem as the (constructiveor destructive) weighted- $ -protection problem, where “weighted” indicates thatthe voters are weighted and “$” indicates that arbitrary awarding and bribingprices are involved. In addition to investigating the general weighted- $ -protection problem, we also investigate the following special cases of it: – the $ -protection problem with w j = 1 for each j (i.e., the voters are notweighted); – the weighted-protection problem with p aj = p bj = 1 for each j (i.e., voters areassociated with the unit awarding price and the unit bribing price); the unit-protection problem with w j = p aj = p bj = 1 for each j (i.e., votersare not weighted, and are associated with the unit awarding price and theunit bribing price). – the symmetric protection problem with p aj = p bj for each j (i.e., the awardingprice and the bribing price are always the same), while noting that differentvoters may have different prices. The goal of this subsection is to prove the following theorem.
Theorem 1.
For any non-trivial scoring rule, both the constructive and de-structive weighted-$-protection problem , is Σ p -complete. The theorem follows from Lemma 1 and Lemma 2 below, which shows the Σ p membership and Σ p -hardness, respectively. Lemma 1.
For any non-trivial scoring rule, both the constructive and destruc-tive weighted- $ -protection problems are in Σ p . Lemma 2.
For any non-trivial scoring rule, both the constructive and destruc-tive weighted- $ -protection problems are both Σ p -hard even if there are only m = 2 candidates.Proof sketch. The proof of Lemma 2 follows from De-Negre (DNeg) variant ofbi-level knapsack problem, which is proved to be Σ p -hard by Caprara et al. [3].We give a brief explanation. In this De-Negre variant, there are an adversaryand a packer. The adversary has a reserving budget ¯ F and the packer has apacking budget ¯ B . There is a set of n items, each having a price ¯ p aj to theadversary, a price ¯ p bj to the packer, and a weight ¯ w j = ¯ p bj to both the adversaryand the packer. The adversary first reserves a subset of items whose total priceis no more than ¯ F . Then the packer solves the knapsack problem with respectto the remaining items that are not reserved; i.e., the packer will select a subsetof the remaining items whose total price is no more than ¯ B but their totalweight is maximized. The De-Negre variant asks if the adversary can reserve aproper subset of items such that the total weight of the unreserved items thatare selected by the packer is no more than some parameter W . The De-Negrevariant is similar to the weighted-$-protection protection problem, because wecan view the defender and attacker in the protection problem respectively asthe adversary and packer in the bi-level knapsack problem. In the case of asingle-winner election with m = 2 candidates, the goal of the defender is toassure that the constructive attacker cannot make the loser get a strictly higherscore than the winner by bribing. This is essentially the same as ensuring thatthe constructive attacker cannot bribe a subset of non-awarded voters whosetotal weight is higher than a certain threshold, which is the same as the bi-levelknapsack problem. ⊓⊔ .2 The Weighted-Protection Problem This is a special case of the weighted-$-protection problem when p aj = p bj = 1.The following theorem used by Faliszewski et al. [8] was originally provedfor another problem. In our context, F = 0 and thus V F = ∅ , it is NP-hard todecide if the constructive attacker can succeed or, equivalently, if the defender cannot succeed. Hence, it is coNP-hard to decide if the defender can succeed andTheorem 2 follows. Theorem 2. (By Faliszewski et al. [8]) If m is a constant, the constructiveweighted-protection problem is coNP-hard for any scoring rule that α , α , . . . , α m are not all equal (i.e., it does not hold that α = α = . . . = α m ). In contrast, the destructive version is easy. Using the fact that m , the number ofcandidates, is a constant, we can prove the following Theorem 3 through suitableenumerations. Theorem 3. If m is a constant, then the destructive weighted-protection prob-lem is in P for any scoring rule. This is the special case of the protection problem with w j = 1 for every j . Thefollowing two theorems illustrate the significant difference (in terms of complex-ity) between the general problem and its special case with symmetricity (i.e., p aj = p bj ). Theorem 4.
For constant m and any non-trivial scoring rule, both the con-structive and destructive $ -protection problems are NP-complete. Theorem 5.
For constant m , both destructive and constructive symmetric $ -protection problems are in P for any scoring rule. The following Theorem 6 shows Σ p -hardness for the most special cases of theconstructive weighted-$-protection problem, namely w j = p aj = p bj = 1 (unit-protection). It thus implies readily the Σ p -hardness for the more general con-structive $-protection and constructive weighted-protection. Theorem 6.
For arbitrary m , the r -approval constructive unit-protection prob-lem is Σ p -complete. Membership in Σ p follows directly from Lemma 1. To prove Theorem 6, itsuffices to show the following. emma 3. For arbitrary m , the r -approval constructive unit-protection problemis Σ p -hard even if r = 4 . To prove Lemma 3, we reduce from a variant of the ∃∀ ∃∀ ∃∀ ′ and defined below. The classical ∃∀ Σ p -hard proved by Mcloughlin [16]. By leveraging the proof byMcloughlin [16], we can show the Σ p -hardness of the ∃∀ ′ problem. ∃∀ ′ : Given a parameter t , three disjoint sets of elements W , X , Y of thesame cardinality, and two disjoint subsets M and M of W × X × Y suchthat M contains each element of W ∪ X ∪ Y at most once. Does there exista subset U ⊆ M such that | U | = t and for any U ⊆ M , U ∪ U is not a perfect matching (where a perfect matching is a subset of triples in which everyelement of W ∪ X ∪ Y appears exactly once)? Proof (Proof of Lemma 3).
Given an arbitrary instance of ∃∀ ′ , we constructan instance of the constructive unit-protection problem in r -approval election asfollows. Recall that r = 4 and thus every voter votes for 4 candidates.Suppose | W | = | X | = | Y | = n , | M | = m , | M | = m .There are 3 n + 2 key candidates, including: – n key candidates, each corresponding to one distinct element of W ∪ X ∪ Y and we call them element candidates. The score of every element candidateis n + ξ ; – one key candidate called leading candidate, whose total score is n + t + ξ − – one key candidate called designated candidate, whose total score is ξ .Here ξ is some sufficiently large integer, e.g., we can choose ξ = ( m + m ) n .Besides key candidates, there are also many dummy candidates, each of scoreeither 1 or m − t + 1. The number of dummy candidates will be determinedlater.There are m + m ( m − t + 1) key voters, including: – m key voters, each corresponding to a distinct triple in M and we callthem M -voters. For each ( w i , x j , y k ) ∈ M , the corresponding voter votesfor the 3 candidates corresponding to elements w i , x j , y k together with theleading candidate; – m · ( m − t + 1) key voters, each distinct triple in M corresponds to exactly m − t +1 voters and we call them M -voters. For every ( w i , x j , y k ) ∈ M , eachof its m − t + 1 corresponding voters vote for the 3 candidates correspondingto elements w i , x j , y k together with one distinct dummy candidate. Sincethe m − t + 1 voters are identical, we can view them as m − t + 1 copies,i.e., every M -voter has m − t + 1 copies.Besides key voters, there are also sufficiently many dummy voters. Each dummyvoter votes for exactly one key candidate and 3 distinct dummy candidates.Dummy voters and dummy candidates are used to make sure that the scoreof key candidates are exactly as we have described. More precisely, if we onlycount the scores of key candidates contributed by key voters, then the elementcandidate corresponding to z ∈ W ∪ X ∪ Y has a score of d ( z ) = d ( z ) + ( m − + 1) d ( z ) where d i ( z ) is the number of occurrences of z in the triple set M i for i = 1 ,
2, and the leading candidate has a score of m . Hence, there are exactly n + ξ − d ( z ) dummy voters who vote for the element candidate corresponding to z , and n + t + ξ − − m dummy voters who vote for the leading candidate.Overall, we create P z ∈ W ∪ X ∪ Y ( n + ξ − d ( z )) + n + t + ξ − m − P z ∈ W ∪ X ∪ Y ( n + ξ − d ( z )) + 3( n + t + ξ − m −
1) + m dummycandidates.As the leading candidate is the current winner, the constructive unit-protectionproblem asks whether the election can be protected against an attacker attempt-ing to make the designated candidate win. The defense budget is F = m − t andthe attack budget is B = n . In the following we show that the defender succeedsif and only if the given ∃∀ ′ instance admits a feasible solution U . “Yes” Instance of ∃∀ ′ → “Yes” Instance of Constructive Unit-Protection. Suppose the instance of ∃∀ ′ admits a feasible solution U , weshow that the answer for constructive unit-protection problem is “Yes”.Recall that each M -voter corresponds to a distinct triple ( w i , x j , y k ) in M and votes for 4 candidates – the leading candidate and the three candidatescorresponding to w i , x j , y k . We do not award M -voters corresponding to thetriples in U , but award all of the remaining M -voters. The resulting cost isexactly F = m − t . In what follows we show that after awarding voters thisway, the attacker cannot make the designated candidate win.Suppose on the contrary, the attacker can make the designated candidatewin by bribing α ≤ t voters among the M -voters, β ≤ m voters among the M -voters, and γ dummy voters. We claim that the following inequalities hold: α + β + γ ≤ n (1a)4 α + 3 β + γ ≥ n + t (1b)Inequality (1a) follows from the fact that the attack budget is n and the attackercan bribe at most n voters. Inequality (1b) holds because of the following. Giventhat a candidate can get at most one score from each voter and that the attackercan bribe at most n voters, bribing voters can make the designated candidateobtain a score at most n + ξ . Hence, the score of each key candidate other thanthe designated one should be at most n + ξ −
1. Recall that without bribery,each of the 3 n element candidate has a score of n + ξ and the leading candidatehas a score of n + t + ξ −
1. Hence, the attacker should decrease at least 1 scorefrom each element candidate and t scores from the leading candidate, leadingto a total score of 3 n + t . Note that an M -voter contributes 1 score to 4 keycandidates, therefore it contributes in total a score of 4 to the key candidates.Similarly an M -voter contributes a score of 3, and a dummy voter contributesa score of 1 to the key candidates. Therefore, by bribing (for example) an M -voter, the total score of all the element candidates and the leading candidatecan decrease by at most 4. Thus, inequality (1b) holds.In the following we derive a contradiction based on Inequalities (1a) and (1b).By plugging γ ≤ n − α − β into Inequality (1b), we have 3 α + 2 β ≥ n + t. Since β ≤ n − α , we have 3 α + 2 β ≤ α + 2 n ≤ n + t . Hence, 3 α + 2 β = α + 2 n = 2 n + t ,nd we have α = t and β = n − t . Note that the defender has awarded every M -voter except the ones corresponding to U , where | U | = t . Hence, every votercorresponding to the triples in U is bribed. Furthermore, as Inequality (1b)is tight, bribing voters makes the designated candidate have a score of n + ξ ,while making each of the other key candidates have a score of n + ξ −
1. Thismeans that the score of each element candidate decreases exactly by 1. Hence,the attacker has selected a subset of M -voters such that together with the M -voters corresponding to triples in U , these voters contribute exactly a score of1 to every element candidate. Let U be the set of triples to which the bribed M -voters correspond, then U ∪ U forms a 3-dimensional matching, which is acontradiction to the fact that U is a feasible solution to the ∃∀ ′ instance.Thus, the attacker cannot make the designated candidate win and the answerfor the constructive unit-protection problem is “Yes”. “No” Instance of ∃∀ ′ → “No” Instance of Constructive Unit-Protection. Suppose for any U ⊆ M , | U | = t there exists U ∈ M such that U ∪ U is a perfect matching, we show that the answer to the constructive unit-protection problem is “No”. Consider an arbitrary set of voters awarded by thedefender. Among the awarded voters, let H be the set of triples that correspondsto the awarded M -voters. As | H | ≤ m − t , | M \ H | ≥ t . We select an arbitrarysubset H ⊆ M \ H such that | H | = t . There exists some H ⊆ M such that H ∪ H is a perfect matching, and we let the attacker bribe the set of voterscorresponding to triples in H ∪ H . Note that this is always possible as every M -voter has m − t + 1 copies, so no matter which M -voters are awarded thebriber can always select one M -voter corresponding to each triple in H . It iseasy to see that by bribing these voters, the score of every element candidatedecreases by 1, and the score of the leading voter decreases by t . Meanwhile, leteach bribed voter vote for the designated candidate and three distinct dummycandidates, then the designated candidate has a score of n + ξ and becomes awinner, i.e., the answer to the constructive unit-protection problem is “No”. ⊓⊔ Remark . The proof of Lemma 3 can be easily modified to prove the Σ p -hardnessof r -approval constructive unit-protection problem for any fixed r ≥
4. Specif-ically, we can make the same reduction, and add dummy candidates such thatevery voter additionally votes for exactly r − Both r -approval destructive weighted-protection and r -approval (sym-metric) $ -protection problems are NP-complete. The proof of Theorem 7 is based on a crucial observation of the equivalence be-tween the destructive weighted-$-protection problem (under an arbitrary scoringrule) and the minmax vector addition problem we introduce (see Appendix 7.1).The full proof of Theorem 7 can be found in Appendix 7.6.
The preceding characterization of the computational complexity of the protec-tion problem in various settings is summarized in Table 1. able 1.
Summary of results for single-winner election under the r -approval scoringrule: “Symmetric” means p aj = p bj for every j and “asymmetric” means otherwise;hardness results that are proved for the case with only two candidates (i.e., m = 2)are marked with a “ ⋄ ” (Note that when m = 2, the 1-approval rule is the same as theplurality, veto or Borda scoring rule. It can be shown that with a slight modification,the hardness results hold for any non-trivial scoring rule); algorithmic results (markedwith a “P”) hold for arbitrary scoring rules; the complexity of the protection problemagainst a destructive attacker with w j = p aj = p bj = 1 remains open; for most variants ofthe protection problem against a constructive attacker, we only provide hardness resultsand we do not know yet whether or not they belong to the class of coNP-complete or Σ p -complete proble. Σ p -complete ⋄ (Thm 1) Σ p -complete ⋄ (Thm 1)Weighted, p aj = p bj = 1 P (Thm 3) coNP-hard (Thm 2) w j = 1, Priced, Asymmetric NP-complete ⋄ (Thm 4) NP-complete ⋄ (Thm 4) w j = 1, Priced, Symmetric P (Thm 5) P (Thm 5) w j = 1, p aj = p bj = 1 P (Thm 5) P (Thm 5)arbitrary Weighted, Priced, Asymmetric Σ p -complete ⋄ (Thm 1) Σ p -complete ⋄ (Thm 1)Weighted, p aj = p bj = 1 NP-complete (Thm 7) Σ p -hard (Thm 6) w j = 1, Priced, Asymmetric NP-complete (Thm 7) Σ p -hard (Thm 6) w j = 1, Priced, Symmetric NP-complete (Thm 7) Σ p -hard (Thm 6) w j = 1, p aj = p bj = 1 ? Σ p -hard (Thm 6) We remark three natural open problems for future research. One is the com-plexity of the destructive protection problem with w j = p aj = p bj = 1. It is notclear whether the problem is in P or is NP-complete. Another is the constructiveprotection problem with p aj = p bj = 1 and arbitrary voter weights. We only showits coNP-hardness, it is not clear whether or not this problem is coNP-complete.The third problem is the complexity of r -approval constructive unit-protectionproblem when r = { , , } as our hardness proof only holds when r ≥ We introduced the protection problem and characterized its computational com-plexity. We showed that the problem, in general, is Σ p -complete, and identifiedsettings in which the problem becomes easier. Moreover, we showed the protec-tion problem in some parameter settings is polynomial-time solvable, suggestingthat these parameter settings can be used for real-work election applications.In addition to the open problems mentioned in Section 5, the following arealso worth investigating. First, our hardness results would motivate the study ofapproximation or FPT (fixed parameter tractable) algorithms for the protectionproblem. Note that even polynomial time approximation schemes can exist for Σ p -hard problems (see, e.g., By Caparara et al. [3]). It is thus desirable thata similar result can be obtained for some variants of the protection problem.Second, how effective is this approach when applied towards the problem ofdefending against other types of attackers that can, e.g., add or delete votes?Third, much research remains to be done in extending the protection problemto accommodate other scoring rules such as Borda and Copeland. Appendix
We provide an equivalent formulation of the destructive weighted-protectionproblem under any scoring rule α = ( α , · · · , α m ), which will be very useful forseveral proofs throughout this paper. The minmax vector addition problem:
Input:
A vector Λ = ( Λ ( c ) , Λ ( c ) , . . . , Λ ( c m − )) where Λ ( c i ) is the score of c i in the absence of bribery. An ( m − ∆ j = ( ∆ j , ∆ j , . . . , ∆ ( m − ,j )for each voter v j where ∆ ij = α − α τ − j ( i ) + α τ − j ( m ) − α m . Awarding price p aj and bribing price p bj for voter v j , j = 1 , , . . . , n . Defense budget F and attackbudget B . Output : Decide if there exists a subset V F ⊆ V such that – P j : v j ∈∈V F p aj ≤ F ; and – For any subset V B ⊆ V \ V F with P j : v j ∈V B p bj ≤ B , it holds that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) Λ + X j : v j ∈V B w j ∆ j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ ≤ Λ ( c m ) , where k·k ∞ is the infinity norm (i.e., the maximal absolute value amongthe m − Lemma 4.
The answer to the destructive weighted- $ -protection problem is “Yes”if and only if the answer to the corresponding minmax vector addition problemis “Yes”.Proof. A “Yes” Instance of Minmax Vector Addition → A “Yes” In-stance of Destructive Weighted-$-Protection.
Suppose the answer to theminmax vector addition problem is “Yes.” Then, there exists some V ∗ F ⊆ V suchthat for any V B ⊆ V \ V ∗ F with P j : v j ∈V B p bj ≤ B , it holds that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) Λ + X j : v j ∈V B w j ∆ j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ ≤ Λ ( c m ) . (2)For showing a contradiction, suppose the answer for the destructive weighted-$-protection problem is “No”. In this case, even if the defender awards the votersin V ∗ F , the attacker can still make c m lose by bribing some subset V ∗ B of voters.Note that if c m does not win, there must exist some other candidate, say, c i , whogets a strictly higher score than c m after the attacker bribes some voters. Let uscompare their scores before and after bribing voters. Before bribing voters, thescores of c i and c m are Λ ( c i ) and Λ ( c m ), respectively. Recall that a candidate c k is at the position of τ − j ( k ) on the preference list of v j , therefore any v j ∈ V ∗ B ontributes a score of α τ − j ( i ) to Λ ( c i ), and contributes a score of α τ − j ( m ) to Λ ( c m ). After bribing voters, the preference list of v j is changed, but regardlessof the change, v j contributes at least α m to c m and at most α to c i . Let thescores of c i and c m after bribing voters be Λ ′ ( c i ) and Λ ′ ( c m ), respectively. Then,it follows that Λ ′ ( c i ) ≤ Λ ( c i ) + X j : v j ∈V ∗ B w j ( α − α τ − j ( i ) ) Λ ′ ( c m ) ≥ Λ ( c m ) + X j : v j ∈V ∗ B w j ( α m − α τ − j ( m )) . Since Λ ′ ( c i ) > Λ ′ ( c m ), we have Λ ( c i ) + X j : v j ∈V ∗ B w j ( α − α τ − j ( i ) ) > Λ ( c m ) + X j : v j ∈V ∗ B w j ( α m − α τ − j ( m ) ) , that is, Λ ( c i ) + X j : v j ∈V ∗ B w j ∆ ij > Λ ( c m ) , 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 destruc-tive weighted-$-protection problem is “Yes” by awarding the voters in V ∗ F . Weshow that the answer to the corresponding instance of minmax vector additionproblem is “Yes”. Suppose on the contrary the answer is “No.” Then, for V ∗ F there exists some V ∗ B ⊆ V \ V ∗ F such that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) Λ + X j : v j ∈V ∗ B w j ∆ j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ > Λ ( c m ) . Consequently, there must exist some 1 ≤ i ≤ m − Λ ( c i )+ P j : v j ∈V ∗ B ∆ ij >Λ ( c m ). By plugging in ∆ ij , we have Λ ( c i ) + X j : v j ∈V ∗ B w j ( α − α τ − j ( i ) ) > Λ ( c m ) + X j : v j ∈V ∗ B w j ( α m − α τ − j ( m ) ) . This means that if the defender awards the voters in V ∗ F , then the attacker canbribe the voters in V ∗ B to change their preference lists such that for any v j ∈ V ∗ B ,candidate c i is on top of the list and c m is at bottom of the list. By doing this, c i gets a strictly higher score than c m . This contradicts the fact that the answerto the destructive weighted-$-protection problem is “Yes”. Hence, the answer tothe minmax vector addition problem is “Yes”. ⊓⊔ .2 Voter Dominance and Preliminary Observations Let S m be the set of permutations over { c , c , . . . , c m } . Each element of S m canbe a preference list. Let V h ⊆ V be the set of voters whose preference list is the h -th element of S m .For two voters v j and v j ′ , we say v j dominates v j ′ (or v j ′ is dominated by v j ), denoted by v j ′ ≺ v j , if any of the following two conditions hold: (i) Thefollowing holds and at least one of the inequalities is strict:( τ j = τ j ′ ) ∧ ( w j ≥ w j ′ ) ∧ ( p aj ≤ p aj ′ ) ∧ ( p bj ≤ p bj ′ ) . (ii) The following holds:( τ j = τ j ′ ) ∧ ( w j = w j ′ ) ∧ ( p aj = p aj ′ ) ∧ ( p bj = p bj ′ ) ∧ ( j ′ < j ) . Note that the domination relation is only defined between the voters who havethe same preference. Intuitively, if v j ′ ≺ v j , then v j is more “important” than v j ′ because v j has a greater weight but is “cheaper” to bribe or award (i.e., morevaluable to both the attacker and the defender).We have the following lemmas. Lemma 5.
Consider the destructive weighted- $ -protection problem with V F ⊆ V being the set of awarded voters. Suppose the attacker can succeed by bribing asubset V B ⊆ V \ V F of voters. If v j ′ ≺ v j , v j ′ ∈ V B and v j ( V F ∪ V B ) , then theattacker can succeed by bribing ( V B \ { v j ′ } ) ∪ { v j } . Towards the proof, we need Lemma 4, which states that the destructiveweighted-$-protection problem is equivalent to another problem called minmaxvector addition. The reader may refer to Section 4.2 for the definition of thisproblem. The following observation follows directly from the definition of ∆ ij which is included in the definition of minmax vector addition. Observation 1 If v j ′ ≺ v j , then ∆ ij ′ ≤ ∆ ij for ≤ i ≤ m − . Assuming Lemma 4, we can prove Lemma 5 as follows.
Proof (Proof of Lemma 5).
We prove the lemma by applying an exchange ar-gument to the minmax vector addition problem, which is equivalent to the de-structive weighted-$-protection by Lemma 4. Suppose by bribing voters in V B the destructive attacker can make c m lose. Then, it follows that P j : v j ∈V B p bj ≤ B and || Λ + X j : j ∈V B w j ∆ j || ∞ > Λ ( c m ) . As v j dominates v j ′ , ∆ ij ≥ ∆ ij ′ and w j ≤ w j ′ . Hence, X j : j ∈ ( V B \{ v j ′ } ) ∪{ v j } p bj ≤ B nd || Λ + X j : j ∈ ( V B \{ v j ′ } ) ∪{ v j } w j ∆ j || ∞ > Λ ( c m ) . That is, the briber can also win by bribing voters in ( V B \ { v j ′ } ) ∪ { v j } . ⊓⊔ Lemma 6.
Consider the destructive weighted- $ -protection problem. Suppose thedefender succeeds by awarding a subset V F ⊆ V of voters. If v j ′ ≺ v j , v j ′ ∈ V F and v j
6∈ V F , then the defender can succeed by awarding ( V F \ { v j ′ } ) ∪ { v j } .Proof. We again use an exchange argument to the minmax vector addition prob-lem. Let V A = V F \ { v j ′ } . Suppose on the contrary that the defender cannot winby fixing voters in V A ∪ { v j } , then there exists some V B ⊆ V \ ( V A ∪ { v j } ) suchthat P j : v j ∈V B p bj ≤ B and || Λ + X j : j ∈V B w j ∆ j || ∞ > Λ ( c m ) . We argue that the defender cannot win either by fixing voters in V A ∪{ v j ′ } , whichis a contradiction. Suppose the defender fixes voters in V A ∪ { v j ′ } . There are twopossibilities. If v j ′
6∈ V B , then we let the briber bribe voters in V B . It is obviousthat the briber can win. Otherwise, v j ′ ∈ V B , then we let the briber bribe votersin ( V B \ { v j ′ } ) ∪ { v j } . Since v j dominates v j ′ , we have P j : v j ∈ ( V B \{ v j ′ } ) ∪{ v j } p bj ≤ B and || Λ + X j : j ∈ ( V B \{ v j ′ } ) ∪{ v j } w j ∆ j || ∞ > Λ ( c m ) . Hence, the lemma is true. ⊓⊔ We say ¯
V ⊆ V is maximal (with respect to V ) if for any ¯ v ∈ ¯ V , there is no v ∈ V \ ¯ V that can dominate ¯ v . That is, ¯ V contains the most important voters.The following corollary follows directly from the preceding two lemmas. Corollary 1
Consider the destructive weighted- $ -protection problem. Withoutloss of generality, we can assume that V F is maximal with respect to V and V B is maximal with respect to V \ V F . Unfortunately, Corollary 1 does not hold for the constructive weighted-$-protection problem, which is significantly different from the destructive versionof the protection problem in terms of computational complexity. Nevertheless,similar results hold for the unweighted constructive problem.
Lemma 7.
Given V F ⊆ V as the set of fixed voters in the constructive $ -bribery-protection problem, suppose a briber can make c i win by bribing a subset V B ⊆V \ V F of voters. If v j ′ ≺ v j , v j ′ ∈ V B and v j ( V F ∪ V B ) , then the briber canalso win by bribing voters in ( V B \ { v j ′ } ) ∪ { v j } .roof. Again, we prove by an exchange argument. Suppose by bribing votersin V B the constructive briber can make c i win. Now we consider the followingprocedure: we change the preference list of v j into the same one as that of v j ′ , and meanwhile, restore the preference list of v j ′ to the original one. Asvoters have the same weight, this procedure does not change the total score ofevery candidate, and c i is thus still the winner. Furthermore, this procedure isequivalent as we bribe ( V B \ { v j ′ } ) ∪ { v j } . Since v j dominates v j ′ , the total costof bribing ( V B \ { v j ′ } ) ∪ { v j } is no more than that of bribing V B . Hence, thelemma is true. ⊓⊔ Lemma 8.
In the constructive $ -bribery-protection problem, suppose the de-fender can win by fixing a subset V F ⊆ V of voters. If v j ′ ≺ v j , v j ′ ∈ V F and v j
6∈ V F , then the defender can also win by fixing voters in ( V F \ { v j ′ } ) ∪ { v j } .Proof. Suppose on the contrary that the defender cannot win by fixing voters in V ′ F = ( V F \ { v j ′ } ) ∪ { v j } . Then the constructive briber can win by bribing votersin some subset V B ⊆ V \ V ′ F . There are two possibilities. If v j ′
6∈ V B , then evenif the defender fixes V F the briber can still bribe V B and make c i win, which isa contradiction. Otherwise v j ′ ∈ V B . If the defender fixes V F , we let the briberbribe ( V B \ { v j ′ } ) ∪{ v j } . According to Lemma 7, if the briber can win by bribing V B , he/she can also win by bribing ( V B \ { v j ′ } ) ∪ { v j } , again contradicting thefact that the defender can succeed by awarding V F . ⊓⊔ The above Lemmas implies the following.
Corollary 2
Without loss of generality, we can assume that V F is maximalwith respect to V , and V B is maximal with respect to V \ V F in the constructive $ -bribery-protection problem. Recall that our goal is to prove Theorem 1.
Theorem 1.
For any non-trivial scoring rule, both the constructive and de-structive weighted-$-protection problem , is Σ p -complete. We first show Σ p -membership. Lemma 1.
For any non-trivial scoring rule, both the constructive and destruc-tive weighted- $ -protection problems are in Σ p . For ease of proof, we use the following definition of Σ p from [19] (see Theorem3 therein). Definition 1 (By Wrathall [19])
Let Γ be a finite set of symbols (alphabet) and Γ + be the set of strings of symbols in Γ . Let L ⊆ Γ + be a language. L ∈ Σ p ifand only if there exists polynomials φ , φ and a language L ′ ∈ P = Σ p suchthat for all x ∈ Γ + , x ∈ L if and only if ( ∃ y ) φ ( ∀ y ) φ [( x, y , y ) ∈ L ′ ] , here ( ∃ y ) φ ( ∀ y ) φ [( x, y , y ) ∈ L ′ ] denotes ( ∃ y )( ∀ y )[ | y | ≤ φ ( | x | ) and if | y | ≤ φ ( | x | ) , ( x, y , y ) ∈ L ′ ] . Proof (Proof of Lemma 1).
Given an instance I of the constructive or destructiveweighted-$-protection problem, we want to know if there exists a subset V F suchthat P j : v j ∈V F p aj ≤ F and for any subset V B ⊆ V \V F with P j : v j ∈V B p bj ≤ B andany preference list ˆ τ j for v j ∈ V B , the following property R ( I, V F , V B ∪ { ˆ τ j | v j ∈V B } ) is true: By bribing voter in V B and change the preference list of each v j ∈V B to ˆ τ j , a constructive attacker cannot make candidate c i win, or a destructiveattacker cannot make c m lose. It is easy to see that the property R ( I, V F , V B ∪{ ˆ τ j | v j ∈ V B } ) can be verified in polynomial time, therefore Lemma 1 is proved. ⊓⊔ Now we prove the Σ p -hardness. Lemma 2.
For any non-trivial scoring rule, both the constructive and destruc-tive weighted- $ -protection problems are both Σ p -hard even if there are only m = 2 candidates. Note that in case of m = 2, destructive weighted-$-bribery-protection isequivalent to constructive weighted-$-bribery-protection. We prove the Lemma 2for the protection problem under plurality. With slight modification the proofworks for any non-trivial scoring protocol for two candidates.We reduce from the De-Negre (DNeg) variant of bi-level knapsack problem,which is proved to be Σ p -hard by Caprara et al. [3]. Before we describe thebi-level knapsack problem, we first introduce the classical knapsack problem,which is closely related. In the knapsack problem, given is some fixed budget¯ B together with a set S of items, each having a price ¯ p aj and a weight ¯ w j . Thegoal is to select a subset of items whose total price is no more than the givenbudget and the total weight is maximized. We denote by KP ( S, ¯ B ) the optimalobjective value of the knapsack problem.In the De-Negre (DNeg) variant of bi-level knapsack problem, there is anadversary and a packer. The adversary has a reserving budget ¯ F and the packerhas a packing budget ¯ B . There is a set of n items, each having a price ¯ p aj to theadversary, ¯ p bj to the packer and a weight ¯ w j = ¯ p bj (to both the adversary andthe packer). The adversary first reserves a subset of items whose total prices isno more than ¯ F . Then the packer solves the knapsack problem with respect tothe remaining items that are not reserved, i.e., the packer will select a subsetof remaining items whose total price is no more than ¯ B such that their totalweight is maximized. The DNeg variant of the bi-level knapsack problem asksfor a proper subset of items reserved by the adversary such that the total weightof items selected by the packer is minimized. More precisely, the problem can beformulated as a bi-level integer programming as follows. he DNeg variant of bi-level knapsack problem: Minimize n X j =1 ¯ p bj y j s.t. n X j =1 ¯ p aj x j ≤ ¯ F where y , y , · · · , y n solves the following:Maximize n X j =1 ¯ p bj y j s.t. n X j =1 ¯ p bj y j ≤ ¯ Bx j + y j ≤ ≤ j ≤ nx j , y j ∈ { , } ≤ j ≤ n The decision version of the DNeg variant of bi-level knapsack problem askswhether there exists a feasible solution with the objective value at most W . Thefollowing lemma is due to Caprara et al. [3]. Lemma 9 (By Caprara et al. [3]).
The decision version of the DNeg variantof bi-level knapsack problem is Σ p -complete. Based on the above lemma, we are able to prove Lemma 2.
Proof (Proof of Lemma 2).
Given an arbitrary instance of the (decision versionof) DNeg variant of bi-level knapsack problem, we construct an election instanceas follows. There are m = 2 candidates. The defense and attack budgets are F = ¯ F and B = ¯ B , respectively. There are n + 2 voters: – n key voters v , v , · · · , v n who vote for c , each having an awarding price p aj = ¯ p aj , a bribing price p bj = ¯ p bj , and a weight w j = ¯ p bj . – one dummy voter v n +1 who votes for c whose weight is 2 W , awarding priceis F + 1 and bribing price is B + 1. – one dummy voter v n +2 who votes for c whose weight is P nj =1 ¯ p bj , awardingprice is F + 1 and bribing price is B + 1.Obviously c is the original winner. We show in the following that the con-structed election instance is secure if and only if the DNeg variant of bi-levelknapsack problem admits a feasible solution with an objective value at most W .Suppose the DNeg variant of bi-level knapsack problem admits a feasiblesolution with an objective value at most W , and let x ∗ i be such a solution. As P nj =1 ¯ p aj x ∗ j = P nj =1 p aj x ∗ j ≤ ¯ F = F , we let the defender award all the voters suchthat x ∗ j = 1, i.e., let V F = { v j | x ∗ j = 1 } . According to the fact that the objectivevalue of the bi-level knapsack problem is at most W , and the fact that the twoummy voters can never be bribed, it follows that the optimal objective valueof the following knapsack problem is at most W :Maximize X j : v j ∈V\V F p bj y j s.t. X j ∈V\V F p bj y j ≤ B Thus, with a budget of B the the briber can never bribe key voters whose totalweight is more than W . Note that originally the total weight of voters voting for c is 2 W + P nj =1 p bj , and the total weight of voters voting for c is P nj =1 p bj , thebriber cannot succeed and the election is thus secure.Suppose the election is secure. Then there exists some V F such that wecan say P j : v j ∈V F p aj ≤ F and if voters in V F are fixed, no briber can succeed.Note that the two dummy voters can never be protected nor bribed, whereas V F ⊆ { v , v , · · · , v n } . Thus, among voters in { v , v , · · · , v n } \ V F , within abudget of B the briber cannot bribe voters whose total weight is more than W .This is equivalent as saying that by setting x j = 1 for v j ∈ V F and x j = 0otherwise, the knapsack problem for y j does not admit a feasible solution withan objective value more than W . Hence, the objective value of the given DNegvariant of bi-level knapsack problem is at most W . ⊓⊔ If m is a constant, then the destructive weighted-protection prob-lem is in P for any scoring rule.Proof. The theorem is proved by trying all different possible V F and checkwhether the attacker can succeed for each of them. By Corollary 1, for vot-ers having the same preference, V F contains voters of the largest weights. Henceto determine V F , it suffices to know the number of voters having each preferencein V F . There are at most m ! different preferences, and consequently at most n m ! different kinds of V F , which is polynomial when m is constant. For each possiblechoice of V F , we check whether the attacker can succeed by trying all possible V B and all possible ways of changing their preferences. Firstly, by Corollary 1,for voters in V \ V F that have the same preference list, V B contains the ones ofthe largest weights, hence using a similar argument we know there are at most n m ! different kinds of V B . Given V F and V B , it remains to determine how thepreference lists of voters in V B should be changed. Note that we do not needto specify how the preference list is changed for each v j ∈ V B . Instead, we onlyneed to determine the number of voters in V B that are changed to each prefer-ence list, which gives rise to at most n m ! possibilities. Therefore, overall thereare at most n m ! different possibilities regarding V F , V B and how to alter thepreference lists of voters in V B , which can be enumerated efficiently when m isa constant. ⊓⊔ We remark that an argument similar to the one we used to prove Theorem 3was used by Faliszewsk et al. [8]. .5 Proofs omitted in Section 3.3Theorem 4.
For any non-trivial scoring rule, both the constructive and destruc-tive $ -protection problems are NP-complete. To prove Theorem 4, we first show the problem belongs to NP in Lemma 10and then show its NP-hardness in Lemma 11.
Lemma 10.
For any scoring rule and arbitrary constant m , both the construc-tive and destructive $ -protection problems are in NP.Proof. Note that to show the membership in NP, it suffices to show that given V F , we can determine in polynomial time whether the constructive/destructiveattacker can succeed. Note that among voters of the same preference list in V \ V F , V B always contains the ones with the smallest bribing prices. Hence,similarly as the proof of Theorem 3, there are at most n m ! different kinds of V B . Given V B , a similar argument as that of Theorem 3 shows that there are n m ! different ways of altering the preference lists of voters in V B . Hence in n m ! time we can determine the whether the constructive/destructive attacker cansucceed, which is polynomial if m is a constant. ⊓⊔ Lemma 11.
For any non-trivial scoring rule, both the constructive and destruc-tive $ -protection problems are NP-hard even if there are only 2 candidates. Again, in case of two candidates, the constructive and destructive variants areidentical and it suffices to prove the theorem under the scoring rule of plurality.Towards the proof, we need the following intermediate problems.
Balanced Partition:
Given a set of positive integers { a , a , · · · , a n } where a ≤ a ≤ · · · ≤ a n and an integer q such that P nj =1 a j = 2 q . Determinewhether there exists a subset S of n integers such that P i : a i ∈ S a i = q .The balanced partition problem is a variant of the partition problem (inwhich S is not required to contain exactly n integers). The NP-completeness ofthe balanced partition problem is a folklore result, which follows from a slightmodification on NP-completeness proof for the partition problem given by Gareyand Johnson[11].Using the balanced partition problem, we are able to show the NP-hardnessof the following problem in Lemma 12. Balanced partition ′ : Given a set of positive integers { a , a , · · · , a n } where a ≤ a ≤ · · · ≤ a n and an integer q such that P nj =1 a j = 2 q , a n + a n +1 + · · · + a n − ≥ q + 1. Determine whether there exists a subset S of n integers such that P i : a i ∈ S a i = q . Lemma 12.
Balanced partition ′ is NP-complete.Proof. Membership in NP is straightforward. We show balanced partition ′ isNP-hard in the following via reduction from balanced partition.Given an instance of the balanced partition problem where the integers are a ≤ a ≤ · · · ≤ a n and q = 1 / · P nj =1 a j , we construct an instance of thebalanced partition ′ problem by adding 4 n integers, each of value 3 q .e first show that the constructed instance is a feasible instance. Obviouslyevery additional integer is larger than any a j where j ≤ n . Let the additionalintegers be a n +1 , a n +2 , · · · , a n . In the constructed instance there are 2 n ′ = 6 n elements, with the summation of all integers being 2 q + 12 nq . Let q ′ = q + 6 nq .Obviously a n ′ +1 + a n ′ +2 + · · · + a n ′ − = 9 nq > q ′ + 1. Thus, the constructedinstance is a feasible instance of the balanced partition ′ problem.We show that the constructed instance admits a feasible partition if and onlyif the given balanced partition instance admits a feasible solution.If the given balanced partition instance admits a feasible solution, then ob-viously the constructed balanced partition ′ problem admits a feasible solution(by adding 2 n of the additional integers to both sides).Suppose the constructed balanced partition ′ instance admits a feasible so-lution. We claim that among the 4 n additional integers, there are exactly 2 n of them in S . Otherwise S contains either at most 2 n − n + 1 of them. By symmetry we assume without loss of generality that S contains at most 2 n − S add up to at most(2 n − · q + 2 q < q ′ = q + 6 nq , which is a contradiction. Thus, S containsexactly 2 n additional integers, implying that the remaining n integers adding upto q , i.e., the given balanced partition instance admits a feasible solution. ⊓⊔ Proof (Proof of Lemma 11).
We will prove the NP-hardness for an election withonly two candidates under 1-approval (plurality). Recall that in this case theconstructive and destructive protection problem are the same.We reduce from the balanced partition ′ problem. Given an arbitrary instanceof the balanced partition ′ problem, we construct an instance of the $-bribery-protection problem such that the answer to the problem is “Yes” if and only thebalanced partition ′ instance admits a feasible solution.We construct the $-bribery-protection instance as follows. There are m = 2candidates. c is the designated candidate. Let F = 4 qn + q , B = (4 n − q − n − – n key voters voting for c , whose awarding prices are 4 q + a , q + a , · · · , q + a n and bribing prices are 4 q − a , q − a , · · · , q − a n , respectively. – n − c , whose awarding prices are all F + 1 andbribing prices are all B + 1. – n dummy voters voting for c , whose awarding and bribing prices are all 1.Let V ∗ = { v , v , · · · , v n } be the set of key voters.Obviously c is the original winner. We show that the answer to the con-structed $-bribery-protection instance is “Yes” if and only if the balanced partition ′ problem admits a feasible solution.Suppose the given balanced partition ′ instance admits a feasible solution S .Now we let the defender fix the n key voters whose awarding price is 4 q + a j where a j ∈ S . It is easy to verify that the total awarding price is 4 nq + q = F ,which stays within the defense budget. We argue that, no briber can alter theelection result with a budget of B . Let V F be the set of fixed voters. Suppose onthe contrary there is a briber who can make c win. Given that the total attackudget is B , the briber can only bribe key voters. Furthermore, the briber has tobribe at least n voters. As |V F | = n , the briber has to bribe all voters in V ∗ \ V F .However, X j : v j ∈V ∗ \V F p bj = (4 n − q > B, which is a contradiction. Thus, the answer to the constructed $-bribery-protectioninstance is “Yes”.Suppose the answer to the constructed $-bribery-protection instance is “Yes”.Note that in total there are 4 n − c and 2 n voters voting for c . Thus any briber who wants to alter the election result has to bribe at least n voters who originally vote for c . Since the 2 n − c can never be protected nor bribed, we can fix a subset V F ⊆ V ∗ such that nobriber can bribe n or more voters from V ∗ \ V F with a budget of B . We have thefollowing claim. Claim. |V F | = n . Proof (Proof of Claim 7.5).
We first show that |V F | ≤ n . Suppose on the contrarythat |V F | ≥ n + 1. Note that any key voter has an awarding price of at least4 q , thus the total awarding price is at least 4( n + 1) q . However, F = 4 qn + q < n + 1) q , which is a contradiction.We now show that |V F | ≥ n . Suppose on the contrary that |V F | ≤ n − n + 1 key voters that can be bribed and we bribe thecheapest n voters. As p b ≥ p b ≥ · · · ≥ p b n , the total bribing price of the cheapest n voters among any n + 1 voters is at most p bn + p bn +1 + · · · + p b n − = 4 qn − ( a n + a n +1 + · · · + a n − ) ≤ qn − ( q + 1) = B , whereas the briber can alwaysbribe the cheapest n voters and let c win, which contradicts the fact that theanswer to the $-bribery-protection instance is “Yes”. Hence |V F | ≥ n . ⊓⊔ Now the following inequalities hold simultaneously: X j : v j ∈V F p aj ≤ F = 4 qn + q (5a) X j : v i ∈V ∗ \V F p bj ≥ B + 1 = 4( n − / q + q (5b)Note that P j : v i ∈V ∗ \V F p bj = P nj =1 p bj − P j : v j ∈V F p bj = 4(2 n − / q − P j : v j ∈V F (4 q − a j ) , by (5a) we have X j : v j ∈V F (4 q − a j ) ≤ qn − q. sing the fact that |V F | = n , we have X j : v j ∈V F a j ≥ q. From (5b), we have X j : v j ∈V F a j ≤ q. Thus, X j : v j ∈V F a j = q, i.e., the given balanced partition ′ instance admits a feasible solution. ⊓⊔ The symmetric $-protection problem (i.e., p aj = p bj ), however, is significantlyeasier, as shown by Theorem 5. Theorem 5.
For constant m , both destructive and constructive symmetric $ -protection problems are in P for any scoring rule.Proof. The proof idea is the same as that of Theorem 3, namely by trying alldifferent possible V F . For each V F , we try all possible V B ’s and all possible waysof altering the preference of voters in V B . Note that every voter has the sameweight and satisfies that p aj = p bj . Therefore, a voter with a smaller (awarding andbribing) price always dominates a voter with a larger price. By Corollary 1 andCorollary 2, for the voters having the same preference, V F contains the votersthat have the smallest prices. Therefore, in order to determine V F , it suffices toknow the number of voters that have the same preference, implying that thereare at most n m ! different kinds of V F ’s. Similarly, given a V F , there are at most n m ! different kinds of V B ’s. Using the same argument as that of Theorem 3, forevery V F and V B , there are at most n m ! ways of altering the preferences of thevoters in V B . Therefore, there are n m ! different possibilities in total, which canbe enumerated efficiently when m is a constant. ⊓⊔ The goal is to prove Theorem 7.
Theorem 7.
Both r -approval destructive weighted-protection and r -approval (sym-metric) $ -protection problems are NP-complete. Towards the proof, we first show the NP-hardness.
Lemma 13.
The r -approval destructive weighted-protection problem is NP-hardfor any r ≥ .roof. We prove the lemma for r = 3. The case of r > r − d = O (1) times in the given triples, which is also known to be NP-hard stated by Kann [13]. Given a 3DM instance with 3 ζ = | W ∪ X ∪ Y | elementsand η = | M | triples such that every element appears at most d = O (1) timesin M , we construct an instance of the destructive weighted bribery-protectionproblem as follows. Here we further require that η ≥ ζ + 2 d . The assumption iswithout loss of generality since if η ≤ ζ + 2 d −
1, then there are at most O (1)triples outside a perfect matching, and the existence of a perfect matching canbe determined by brute-forcing within ζ O (1) time, which is polynomial.For ease of description, we re-index all elements of W ∪ X ∪ Y arbitrarily as z , z , · · · , z ζ .Let Q = 2 η + 1. There are 3 ζ + 1 key candidates, including the following twokinds of candidates (the function f will be defined later): – ζ key candidates c to c ζ , with c i corresponding to z i ∈ W ∪ X ∪ Y andhas a score of Q · f ( z i ). We call them element candidates; – one key candidate c ζ +1 called leading candidate, which is the original winnerand has a score of Q · f ( z ζ +1 ).Besides key candidates, there are also sufficiently many dummy candidates c i for i > ζ + 1, each having a score of 1. The number of dummy candidates willbe determined later.There are η key voters v to v η , each of weight Q . Each key voter correspondsto a distinct triple in ( z i , z j , z k ) ∈ M , and votes for the three candidates thatcorrespond to z i , z j , z k , respectively.Besides key voters, there are also sufficiently many dummy voters v j for j > η . A dummy vote has a unit weight, and votes for one key candidate andtwo distinct dummy candidates.Now we determine the number of dummy voters and dummy candidatestogether with all the parameters. If we only consider key voters of weight Q ,then every element candidate corresponding to some z i gets a score of Q · d ( z i )where d ( z i ) is the number of occurrences of z in M . Adopting the viewpoint ofthe minmax vector addition problem, for every 1 ≤ j ≤ η we have ∆ ij = , if v j votes for c i and does not vote for c ζ +1 , if v j votes for c ζ +1 and does not votes for c i , or v j votes for both c i and c ζ +1 , if v j votes for c ζ +1 and does not vote for c i Let ∆ max = max ≤ i ≤ ζ ζ X j =1 ∆ ij , d max = max ≤ i ≤ ζ d ( z i ) . e define f ( z i ) = 2 η + d max + ∆ max − η X j =1 ∆ ij , ≤ i ≤ ζ That means, candidate c i , 1 ≤ i ≤ ζ will get a score of Q · d ( z i ) from key voters,and additionally Q · [ f ( z i ) − d ( z i )] ≥ c i we need to create Q · [ f ( z i ) − d ( z i )] dummy voters.We define f ( z ζ +1 ) = 2 η + d max + ∆ max − ζ + 1 . Note that P ηj =1 ∆ ij ≥ η − d > ζ as every element appears at most d times intriples, hence f ( z ζ +1 ) > f ( z i ) for 1 ≤ i ≤ ζ and c ζ +1 is indeed the originalwinner.Overall, dummy voters should contribute Q · [ f ( z i ) − d ( z i )] to each c i , 1 ≤ i ≤ ζ and ( ζ + 1) f ( z ζ +1 ) to c ζ +1 . We create in total Q · [ P ζ +1 i =1 f ( z i ) − P ζi =1 d ( z i )]dummy voters, and 2 Q · [ P ζ +1 i =1 f ( z i ) − P ζi =1 d ( z i )] dummy candidates.Let the defense budget be F = ζ and the attack budget be B = η − ζ . “Yes” Instance of 3DM → “Yes” Instance of Destructive Weighted-Bribery-Protection. Suppose the given 3DM instance admits a feasible solu-tion, we show that the answer to destructive weighted-bribery-protection prob-lem is “Yes”. Let T ⊆ M be the perfect matching. Then | T | = ζ and we let thedefender protect voters corresponding to the triples in T . Taking the viewpointof the minmax vector addition problem. If the attacker bribes all the key voters,then W ( c i ) increases by exactly Q · P ηj =1 ∆ ij for 1 ≤ i ≤ ζ . First it is easyto see that no dummy candidate can be a winner as Q · P ηj =1 ∆ ij ≤ ηQ while Q · f ( z ζ +1 ) ≥ (2 η + 1) Q . Meanwhile, for each key candidate c i , 1 ≤ i ≤ ζ ,his/her total score becomes exactly Q · [ f ( z ζ +1 ) + ζ − Q ( ζ −
1) to each c i , hence after bribery every keycandidate has a score at most Q · f ( z ζ +1 ), implying that the answer to theDestructive Weighted-Bribery-Protection problem is “Yes”. “No” Instance of 3DM → “No” Instance of Destructive Weighted-Bribery-Protection. Suppose the given 3DM instance does not admit a perfectmatching, we show that the answer to the constructed instance of the destructiveweighted-bribery-protection problem is “No”. Consider an arbitrary set of votersfixed by the defender and let U be the subset of key voters that are fixed.Obviously | U | ≤ ζ . If | U | < ζ , we add arbitrary key voters into U such that itscardinality becomes ζ . Let U ′ be the set of these ζ key voters and let the attackerbribe the remaining ζ − η key voters. Again we take the viewpoint of the minmaxvector addition problem. If the attacker bribes every key voter, then the totalscore of every key candidate c i , 1 ≤ i ≤ ζ , becomes exactly Q · [ f ( z ζ +1 )+ ζ − U are not bribed, we subtract their contribution from each c i .Note that triples corresponding to voters in U do not form a perfect matching,thus there exists some element which appears at least twice in these triples. Let k be such element and we consider c k . It is clear that ∆ kj = 0 if v j votes for c k and ∆ kj = 1 if v j does not vote for c k (note that key voters never vote for c ζ +1 ). Hence P j : v j ∈ U ∆ kj ≤ ζ −
2. By subtracting the contribution of voters in U , c k has a score at least Q · [ f ( z ζ +1 + 1)], implying that after bribery c k willget a higher score than c ζ +1 . Thus, the answer to the Destructive Weighted-Bribery-Protection problem is “No”. ⊓⊔ Note that in the preceding reduction, we only construct voters of two differentweights, Q for the key voters and 1 for the dummy voters. Recall that Q is setto be large enough to assure that only the key voters will be considered by thedefender or the attacker. Once V F and V B are restricted to be subsets of the keyvoters, the concrete value of Q does not matter. Moreover, we can also provethe NP-hardness of the destructive (symmetric) $-bribery-protection problemby using essentially the same proof, except that we set key voters of price 1 anddummy voters of price exceeding budgets F and B , say, max { F, B } + 1. Thisleads to the following lemma. Lemma 14.
The r -approval destructive (symmetric) $ -bribery-protection prob-lem is NP-hard for any r ≥ . Having showed the NP-hardness of the destructive weighted-protection prob-lem, we show the problem is polynomial-time verifiable and is therefore NP-complete.
Lemma 15.
The destructive weighted-protection problem can be verified in poly-nomial time under any scoring rule.Proof.
We leverage the minmax vector addition problem. In the case of unitprice, given V F , the decision version of the verification problem becomes: doesthere exist a subset V B ⊆ V \ V F such that the following is true (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) Λ + X j : v j ∈V B ∆ j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) > Λ ( c m ) . To answer this decision problem, it suffices to do the following for every 1 ≤ i ≤ m −
1: pick B voters from V \ V F whose i -th coordinate ∆ ij is the largest, addthem to Λ ( c i ), and check if it is greater than Λ ( c m ). ⊓⊔ It is, however, not clear if the destructive $-protection problem is NP-completefor arbitrary scoring rules. However, we show in the following that for any scor-ing rule which only assigns a constant number of different scores to a preferencelist, i.e., the α i ’s only take O (1) distinct values, the $-protection problem can beverified in polynomial-time. As in the case of the r -approval rule, the α i ’s onlytake values of 1 or 0, the destructive $-protection problem is NP-complete forthe r -approval rule. Lemma 16.
The destructive (symmetric) $ -protection problem can be verifiedin polynomial-time in n under any scoring rule in which the α i ’s only take aconstant number of distinct values.roof. Consider the minmax vector addition problem. We observe that in thecase of unit weight and that the α i ’s take O (1) distinct values, ∆ ij only takes O (1) distinct values. For each coordinate i , we can check if it is possible for Λ ( c i ) and ∆ ij ’s to add up to some value strictly greater than Λ ( c m ). Note thatby adding every ∆ ij , we need to pay a price of p bj , hence it is essentially the knapsack problem with items having arbitrary prices but only O (1) distinctweights. Such a knapsack problem can be solved in polynomial-time, e.g., bysimply guessing the number of items of the same weight. Among the items ofthe same weight, the optimal solution should take the ones with the cheapestprice. ⊓⊔ eferences
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