1 Introduction
Democratic societies use elections to select their leaders. However, in societies without a strong democratic tradition, elections may be used as a way to legitimize the status quo: voters are asked to cast their ballots, but the election authorities do not count these ballots correctly, in order to produce an outcome that favors a specific candidate. There are multiple reports of such cases in Russia^{1}^{1}1https://reut.rs/2Gf2FD5, Congo^{2}^{2}2https://on.ft.com/2SW7ggy and Colombia^{3}^{3}3https://colombiareports.com/votingfraudincolombiahowelectionsarerigged/, as well as a number of other countries. Even when the election authorities are trustworthy, election results may be corrupted by an external party, for instance, by means of hacking electronic voting machines (Springall et al., 2014; Halderman and Teague, 2015).
There are several ways to counteract electoral fraud. One approach is to send observers to polling stations, to ensure that only eligible voters participate in the elections and their ballots are counted correctly. However, it may be infeasible for the party that wants to protect the elections (the defender) to send observers to all polling stations. Consequently, the election manipulator (the attacker) may observe which polling stations remain unprotected, and focus their effort on these stations. Thus, under this approach the attacker benefits from the secondmover advantage.
An alternative approach that the defender can explore is to request recounts in some of the voting districts. While recounts cannot protect from all forms of attacks on election integrity (e.g., a recount is of limited use if voters have been bribed to vote in a specific way, or if the polling station has been burned down), they are feasible in a range of settings and offer the defender the secondmover advantage. Indeed, there are several examples where a recount changed the election outcome. For instance, in the 2008 United States Senate election in Minnesota the Democratic candidate Al Franken won the seat after a recount revealed that 953 absentee ballots were wrongly rejected^{4}^{4}4https://bit.ly/2S2PMxY, and in the 2004 race for governor in Washington the Democratic candidate Gregoire was declared the winner after three consecutive recounts^{5}^{5}5https://bit.ly/2tnO4gG.
However, recounts can be costly. In Gregoire’s case, the Democratic party paid $730000 for a statewide manual recount, and in the 2016 US Presidential Election the fee to initiate a recount in Wisconsin was $3.5 million. Thus, a party that would like to initiate a recount in order to rectify the election results should allocate its budget carefully. Of course, the attacker also incurs costs to carry out the fraud: local election officials may need to be bribed or intimidated, and the more districts are corrupted, the higher is the risk that the election results will not be accepted.


Plurality over Voters ()  Plurality over Districts ()  
Unweighted  Weighted  


Rec  NPc, Thm. 3.1 (i) [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]3  P, Thm. 4.3  NPc, Thm. 4.1 (i) [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]3 
NPc, Thm. 3.1 (ii) [colframe=black,colback=gray!50,coltext=black,shrink tight,boxrule=0.5pt,extrude by=0.5mm]U  NPc, Thm. 4.1 (ii) [colframe=black,colback=gray!50,coltext=black,shrink tight,boxrule=0.5pt,extrude by=0.5mm]U  
, Thm. 3.2  , Thm. 4.2  
Man  NPh, Thm. 3.3 (i) [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]3 [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]0 [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]  NPc, Thm. 4.8 [colframe=black,colback=gray!50,coltext=black,shrink tight,boxrule=0.5pt,extrude by=0.5mm]U  c, Thm. 4.6 [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]3 
NPh, Thm. 3.3 (ii) [colframe=black,colback=gray!50,coltext=black,shrink tight,boxrule=0.5pt,extrude by=0.5mm]U [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]0 [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]  NPh, Thm. 4.7 [colframe=black,colback=gray!50,coltext=black,shrink tight,boxrule=0.5pt,extrude by=0.5mm]U [colframe=black,colback=gray!15,shrink tight,boxrule=0.5pt,extrude by=0.5mm]0  

Our Contribution.
In this paper we analyze the strategic game associated with vote recounting. In our model, there are two players: the attacker, who modifies some of the votes in order to make his preferred candidate the election winner, and the defender, who observes the attacker’s actions and tries to restore the correct outcome (or, more broadly, to ensure that a candidate who is better than wins the election) by means of recounting some of the votes. We assume that the set of voters is partitioned into electoral districts, and both the defender and the attacker make their choices at the level of districts rather than individual votes. The attacker selects a subset of at most districts and changes the vote counts in the selected districts, and the defender can then restore the vote counts in at most districts to their original values. We assume that both players have full information about the true votes and each other’s budgets, and the defender can observe the attacker’s actions. While the full information assumption is not entirely realistic, we note that in a districtbased model both parties only need to know the vote counts in each district rather than individual votes, and one can get fairly accurate districtlevel information from independent polls. Also, verifying whether the votes in a district have been tampered with is possible using risklimiting audits Lindeman and Stark (2012); Schürmann (2016).
For simplicity, we focus on the Plurality voting rule, where each voter votes for a single candidate. We consider two implementations of this rule: (1) Plurality over Voters, where districts are only used for the purpose of collecting the ballots and the winner is selected among the candidates that receive the largest number of votes in total, and (2) Plurality over Districts, where each district selects a preferred candidate using the Plurality rule, and the overall winner is chosen among the candidates supported by the largest number of districts; we also consider a variant of the latter rule where districts have weights, and the measure of a candidate’s success is the total weight of districts that support her. Both of these rules are widely used in practice. For example, Plurality over Voters is commonly used in gubernatorial elections in the US, while Plurality over Districts is used in the US Presidential elections.
We provide a detailed analysis of the computational complexity of the algorithmic problems faced by the attacker and the defender. Our main results are summarized in Table 1. Briefly, assuming that the vote counts and the weights of the districts are specified in binary, most of the problems we consider are computationally hard; however, the defender’s problem appears to be easier than that of the attacker, and we also get some tractability results for the former. Towards the end of the paper, we consider a variant of our model where the attacker is limited to only transferring votes to his preferred candidate; we show that, while this assumption reduces the attacker’s ability to achieve his goals, it lowers the complexity of some of the problems we consider.
Related Work.
There is a very substantial literature on voting manipulation and bribery; we point the readers to the excellent surveys of Conitzer and Walsh (2016) and Faliszewski and Rothe (2016). In much of this literature it is assumed that the malicious party can change some of the votes subject to various constraints, and the challenge is to determine whether the attacker’s task is computationally feasible; there is no defender that can counteract the attacker’s actions.
While there is a number of papers that apply gametheoretic analysis to the problem of voting manipulation, they typically consider interactions between several manipulators, with possibly conflicting goals (e.g., see the recent book by Meir (2018)), rather than a manipulator and a sociallyminded actor. An important exception, which is similar in spirit to our paper, is the recent work of Yin et al. (2018), who investigate a preemptive approach to protecting elections. In their model the defender allocates resources to guard some of the electoral districts, so that the votes there cannot be corrupted; notably, in this model the defender has to commit to its strategy first, and the attacker can observe the defender’s actions before deciding on its response. The leaderfollower (defenderattacker) structure of this model is in the spirit of a series of successful applications of Stackelberg games to security resource allocation problems (Tambe, 2011). Li et al. (2017) analyze a variant of the model of Yin et al. where the goal is to minimize resource consumption, and Chen et al. (2018) study a similar scenario, in which manipulation is achieved through bribing the voters. The key difference between our work and the above papers is the action order of the players: in all prior work on election protection that we are aware of the defender makes the first move.
2 The Model
We consider elections over a candidate set , . There are voters who are partitioned into pairwise disjoint districts , ; for each , let . For each , district has a weight , which is a positive integer; we say that an election is unweighted if for all . Each voter votes for a single candidate in . For each and each let denote the number of votes that candidate gets from voters in ; we refer to the list as the vote profile.
Let be a linear order over ; indicates that is favored over . We consider the following two voting rules, which take the vote profile as their input.

Plurality over Voters (PV). We say that a candidate beats a candidate under PV if or and ; the winner is the candidate that beats all other candidates. Note that district weights are not relevant for this rule.

Plurality over Districts (PD). For each the winner in is chosen from the set , with ties broken according to . Then, for each , , we set if and otherwise. We say that a candidate beats a candidate under PD if or and ; the winner is the candidate that beats all other candidates.
For PV and PD, we define the social welfare of a candidate as the total number of votes that gets and the total weight that gets, respectively:
Hence, the winner under each voting rule is a candidate with the maximum social welfare.
We consider scenarios where an election may be manipulated by an attacker, who wants to change the election result in favor of his preferred candidate . The attacker has a budget , which means that he can manipulate at most districts. For each , we are given an integer , , which indicates how many votes the attacker can change in district if he chooses to manipulate it. Formally, a manipulation is described by a set , , and a vote profile such that for all , , and for all it holds that and .
After the attack, a defender with budget can demand a recount in at most districts. Formally, a defender’s strategy is a set with ; after the defender acts, the vote counts in all districts in are restored to their original values, i.e., the resulting vote profile satisfies for each , and for each , . Then the underlying voting rule is applied to with ties broken according to ; let denote the candidate selected in this manner. The defender chooses her strategy so as to maximize , breaking ties using .
We say that the attacker wins if he has a strategy such that, once the defender responds optimally, candidate is the winner in the resulting vote profile ; otherwise we say that the attacker loses. We note that if , the defender can always ensure that , i.e., the winner at is the winner at the original vote profile , so in what follows we assume that the attacker’s strategy satisfies .
Example 2.1.
Consider an election with five districts over a candidate set , where is the attacker’s preferred candidate; suppose that ties are broken according to the priority order . In each of and there are voters who vote for , and in each of , and there are voters who vote for . Suppose that and for each , and , .
If the voting rule is , then the attacker does not have a winning strategy. Indeed, consider an attacker’s strategy . If , the defender can set ; in the recounted vote profile gets at least votes, so it is the election winner. If , the defender can set : in the recounted vote profile gets at most votes, while gets at least votes, so the winner is or ( can win if, e.g., the attacker chooses to transfer exactly votes from to in , in which case gets votes after the recount). Note that even if the winner in is rather than , the defender still prefers recounting to no recounting: even though she cannot restore the correct result, she prefers to , since .
If the voting rule is , then the attacker can win by choosing and transferring a majority of votes from to in both districts. Indeed, even if the defender demands a recount in one of these districts, still wins the remaining district, leading to a vote weight of in the recounted profile. Since ’s vote weight is and ’s vote weight is , wins by the tiebreaking rule. ∎
We assume that both the defender and the attacker have full information about the game. Both parties know the true vote profile , the parameters and for each district and each others’ budgets. Moreover, the defender observes the strategy of the attacker.
We can now define the following decision problems for each :

Man: Given a vote profile , the attacker’s preferred candidate , budgets and , and district parameters , does the attacker have a winning strategy?

Rec: Given a vote profile , a distorted vote profile with winner , a candidate , a budget , and district weights , can the defender recount the votes in at most districts so that gets elected?
We will also consider an optimization version of Rec, where is not part of the input and the goal is to maximize the social welfare of the eventual winner.
Unless specified otherwise, we assume that the vote counts and the district weights are given in binary; we explicitly indicate which of our hardness results still hold if these numbers are given in unary. All problems considered in this paper admit straightforward greedy algorithms for , so in what follows we focus on the case . When the voting rule is clear from context, we write instead of .
Next, we give formal definitions of the decision problems that are used throughout the paper to show hardness of Rec and Man for , under various constraints.
Definition 2.2 (Subset Sum).
An instance of Subset Sum is given by a multiset of integers. It is a yesinstance if there exists a nonempty subset such that , and a noinstance otherwise.
Definition 2.3 (Exact Cover By 3Sets (X3c)).
An instance of X3C is given by a set of size and a collection of element subsets of . It is a yesinstance if there exists a subcollection of size such that , and a noinstance otherwise.
Definition 2.4 (Independent Set).
An instance of Independent Set is a graph and an integer . It is a yesinstance if there exists a subset of size that forms an independent set, i.e., for all , and a noinstance otherwise.
Definition 2.5 (Partition).
An instance of Partition is given by a multiset of positive integers. It is a yesinstance if there exists a subset such that , and a noinstance otherwise.
All of these problems are NPcomplete (Garey and Johnson, 1979). However, Subset Sum and Partition are NPhard only when the input is given in binary; for unary input, these problems can be solved in time polynomial in the size of the input.
3 Plurality over Voters
In this section we focus on Plurality over Voters. We first take the perspective of the defender, and then the perspective of the attacker.
Unfortunately, the defender’s problem turns out to be computationally hard, even if there are only three candidates or if the input vote counts are given in unary.
Theorem 3.1.
Rec is NPcomplete even when

, or

the input vote profile is given in unary.
Proof.
This problem is clearly in NP. We give separate hardness proofs for the case (part (i)) and for the case where the input is given in unary (part (ii)).
Part (i).
To prove that Rec is NPhard for , we provide a reduction from Subset Sum; see Definition 2.2.
Given an instance of Subset Sum with , we construct an instance of Rec as follows. Without loss of generality, we assume that for every and , and let , , . We set , where is the attacker’s preferred candidate. In what follows, we describe each district by a tuple . There are voters distributed over districts, which are further partitioned into two sets and as follows:

For each there is a district in with votes , which are distorted to , and for each there is a district in with votes , which are distorted to . Note that .

contains three districts with votes , , and , respectively. The votes in these districts are not distorted.
Finally, .
Before the manipulation, gets votes and and get votes each. After the manipulation, gets votes, gets and gets votes; thus, by our assumption that , candidate is the winner in the manipulated profile. The goal is to restore the true winner .
Now, assume that there exists a subset with such that . Then, by recounting the districts of that correspond to the integers in , the defender can ensure that both and get votes. Since always gets votes from the nonmanipulated districts, she is successfully restored as the winner.
Conversely, assume that there is no nonempty subset such that . Then, since the votes of and always add up to exactly , and each of them gets an even number of votes from each district, one of them must get at least votes. Therefore, cannot be restored as the winner.
Part (ii).
We give a reduction from Exact Cover By 3Sets (X3C); see Definition 2.3. Given an instance of X3C, we construct the following Rec instance. Without loss of generality, we assume that , and let .

Let , .

For each subset , there is a district , where gets votes, gets votes, for each candidate gets votes, and for each candidate gets votes. The attacker distorts the votes in by transferring two votes from to each candidate with , so that in the distorted profile gets votes in and every other candidate gets votes in .

There is a district where receives votes, receives votes and for every candidate receives votes; the votes in this district are not distorted.

The budget of the defender is .
Candidate is the true winner with votes, compared to the votes of and the votes of for every . In the distorted profile candidate gets votes, candidate gets votes, and each candidate in gets votes.
Recounting a district reduces by the votes of each candidate such that , leading to getting more votes than these candidates; cannot get more than votes no matter what the defender does. Therefore, can be restored as the winner by recounting districts if and only if can be covered by sets from . ∎
If the number of candidates is bounded by a constant and the input is given in unary, an optimal set of districts to recount can be identified in time polynomial in the input size by means of dynamic programming.
Theorem 3.2.
Rec can be solved in time .
Proof.
Consider an instance of Rec with a candidate set , , and voters that are distributed over districts. For each , let and denote, respectively, the true and distorted votes in district . Let be the budget of the defender.
We present a dynamic programming algorithm that given a candidate , decides whether can be made the election winner by recounting at most districts. Our algorithm fills out a table containing entries of the form , for each , , and ; thus, . We define if we can recount at most of the first districts so that the vote count of candidate equals for each ; otherwise we define . There exists a recounting strategy that restores if and only if there exists a such that , for all , and for all such that the tiebreaking rule favors over .
For each , let be the number of votes that candidate gets after manipulation, and let . We fill out according to the following rule:
This completes the proof. ∎
We obtain similar hardness results for the attacker’s problem. However, it is not clear if Man is in NP. Indeed, it may belong to a higher level of the polynomial hierarchy: it is not hard to see that Man is in , and it is plausible that this problem is hard for this complexity class.
Theorem 3.3.
Man is NPhard even when , for all and

, or

the input vote profile is given in unary.
Proof.
We prove the two claims separately.
Part (i).
To prove that Man is NPhard for , we provide a reduction from Subset Sum; see Definition 2.2.
Given an instance of Subset Sum with , we construct an instance of PVMan as follows. We can assume without loss of generality that and for every , and let ; by our assumptions, . We set , where is the attacker’s preferred candidate. In what follows, we describe each district by a tuple . There are voters distributed over districts, which are further partitioned into four sets as follows:

For each there is a district in with votes . Thus, .

Set consists of districts with votes in each district.

For each there are two districts in with votes . Thus, .

Set consists of three districts with votes , , and .
We set , and for each .
We have and . Hence, the true winner is or , depending on the tiebreaking rule. We claim that the attacker can make the winner if and only if there exists a nonempty subset such that .
To see this, assume first that there exists a subset such that and . Then the attacker can distort the votes in the districts of corresponding to the elements of , and in arbitrary districts of , by transferring all votes to in each of these districts. In the resulting election, gets votes, while and get votes each, so becomes the winner.
Conversely, suppose that the attacker has a successful manipulation with . For each , let denote the number of votes that receives in . For to be the winner in , it must hold that ; since is an integer and , this means that the manipulation transfers at least votes to . On the other hand, in every district there are at most voters who vote for or , so can gain at most votes from the manipulation. It follows that , . If these votes are not split evenly between and , at least one of these candidates would get strictly more than points; since each district allocates an even number of votes to both and , this further means that one of them would get at least votes, a contradiction with being the winner at . Thus, it must be the case that .
Further, , implies that and . Moreover, we have for every district . Hence,
where is the integer in that corresponds to district . Thus, , and hence is a witness that is a yesinstance of Subset Sum.
Part (ii).
To prove that Man is NPhard when the input is given in unary, we provide a reduction from X3C; see Definition 2.3.
Given an instance of X3C with , , we construct an instance of Man as follows. We set , where is the attacker’s preferred candidate. The districts are partitioned into three sets :

For each subset the set contains a district . In this district each candidate such that gets votes, and all other candidates get no votes. Thus, .

For each element , the set contains districts; each of these districts consists of a single voter who votes for .

The set contains a single district that consists of voters who vote for .
We set , and for all .
We have for all and . Hence, the true winner is the candidate in who is favored by the tiebreaking rule. We show that the attacker is able to make the winner if and only if admits an exact cover by sets from .
Suppose that is an exact cover for ; note that . The attacker can manipulate the districts in that correspond to sets in by reassigning all the votes in each of them to . In the resulting election, gets votes, while every other candidate gets votes, as every is covered by exactly one set in .
Conversely, suppose the attacker has a successful manipulation with . For each , let denote the number of votes that receives in . As can gain at most votes for each district in , we have . Let ; note that . We claim that is a cover for . Indeed, if for some no district in is manipulated, the manipulation lowers the score of by at most , so , a contradiction. ∎
In the hardness reductions in the proof of Theorem 3.3 the defender’s budget is
. This indicates that the attacker’s problem remains NPhard even if the defender is known to use a heuristic (e.g., a greedy algorithm) to compute her response.
We remark that Rec and Man with are very similar in spirit to combinatorial (shift) bribery (Bredereck et al., 2016). In both models, a budgetconstrained agent needs to select a set of votechanging actions, with each action affecting a group of voters. However, there are a few technical differences between the models. For instance, in our model different actions are associated with nonoverlapping groups of voters, which is not the case in combinatorial shift bribery. On the other hand, in shift bribery under the Plurality rule votes can only be transferred to/from the manipulator’s preferred candidate , while our model does not impose this constraint (see, however, Section 5). Consequently, it appears that the technical results in our paper cannot be derived from known results for combinatorial shift bribery.
4 Plurality over Districts
In this section we study Plurality over Districts. For the defender’s problem, we can replicate the results we obtain for Plurality over Voters, by using similar techniques.
Theorem 4.1.
Rec is NPcomplete even when

, or

the input vote profile and district weights are given in unary.
Proof.
This problem is clearly in NP. We give separate hardness proofs for the case (part (i)) and for the case where the input is given in unary (part (ii)).
Part (i).
We use the same reduction as in the proof of the first part of Theorem 3.1. An important feature of this reduction is that all voters in each district vote for the same candidate. Thus, if we set the weight of each district to be equal to the number of voters therein, the proof goes through without change.
Part (ii).
We provide a reduction from Independent Set; see Definition 2.4. Given an instance of Independent Set, where , we construct an instance of Rec as follows. Let , ; we can assume without loss of generality that . We set , where is the attacker’s preferred candidate; thus, . We create the following districts. For our argument, the district sizes and the values of do not matter; for concretness, we assume that each district consists of a single voter, whose vote can be changed by the manipulator.

For each edge , there are two districts and with weight each. In each such district the winner before manipulation is , and the winner after manipulation is .

For each node , there is a district with weight ; in this district the winner before manipulation is , and the winner after manipulation is .

There is a set of districts with weight each^{6}^{6}6For convenience, we use fractional weights. We can turn all weight into integers, by multiplying them by .; in each such district the winner before manipulation is , and the winner after manipulation is .

There is a district of weight with winner ; this district is not manipulated.

For each , there is a district of weight with winner ; this district is not manipulated.

For each , there is a district of weight with winner ; this district is not manipulated.
The budget of the defender is . The candidates’ weights before and after manipulation are given in the following table:


true weight  distorted weight  


,  
,  

Hence, the true winner is candidate and the winner after manipulation is .
If is an independent set of size in , the defender can proceed as follows. For each , she demands a recount in and in each district such that is incident to . Since forms an independent set, this requires recounting at most districts. Moreover, after the recount the weight of is , the weight of is , the weight of each candidate such that is , the weight of each candidate such that is at most , and the weight of each candidate such that is at most . Thus, this recounting strategy successfully restores as the election winner.
Conversely, suppose that the defender has a recounting strategy that results in making the election winner. Since , at most districts in can be recounted, so ’s weight after the recount is at most . Now, if contains at most districts in , then ’s weight after the recount is at least , a contradiction with becoming the winner after the recount. Hence, contains at least districts in ; let be the subset of nodes corresponding to these districts. We claim that forms an independent set in .
Indeed, consider a node . If the defender does not recount some district such that is incident to then after the recount the weight of is at least , a contradiction with becoming the winner after the recount. Thus is necessarily recounted. Now, suppose that for some . We have just argued that both and have to be recounted. But this means that the score of is at least after the recount, a contradiction again. Thus, is an independent set. ∎
Theorem 4.2.
Rec can be solved in time .
Proof.
The algorithm is a simple adaptation of the dynamic program presented in the proof of Theorem 3.2. ∎
We also obtain a positive result that does not have an analogue in the PV setting; if all districts have the same weight, the recounting problem can be solved efficiently.
Theorem 4.3.
Rec can be solved in polynomial time if for all .
Proof.
We reduce our problem to nonuniform bribery (Faliszewski, 2008). An instance of nonuniform bribery under the Plurality rule is given by a set of voters and a set of candidates; for each voter and each candidate there is a price for making voter vote for , and the briber’s goal is to make her preferred candidate the Plurality winner^{7}^{7}7Faliszewski (2008) assumes that ties are broken in favor of the briber, but his results extend to lexicographic tiebreaking. while staying within a budget . This problem is known to be in P (Faliszewski, 2008). To reduce Rec to nonuniform bribery, we map each district to a single voter ; if the true winner in is , and in the distorted profile the winner in is , we set , for , and if (i.e., if the attacker has changed the outcome in ), we set . Then for any candidate it holds that in Rec the defender can make win by recounting at most districts if and only if in our instance of nonuniform bribery the briber can make win by spending at most . ∎
We now consider the attacker’s problem. It turns out that for the rule we can obtain a stronger hardness result than for : we will now argue that when weights and vote counts are given in binary, Man is complete even for . Our reduction uses a variant of the Subset Sum problem, which we term SubSubset Sum (SSS); this problem may be of independent interest.
Definition 4.4 (SubSubset Sum).
An instance of SubSubset Sum is a set and a positive integer . It is a yesinstance if there is a subset with such that for every nonempty subset , and a noinstance otherwise.
Our proof proceeds by establishing that SSS is complete (Lemma 4.5; the proof can be found in the appendix), and then reducing this problem to Man.
Lemma 4.5.
SSS is complete.
Theorem 4.6.
Man is complete, even when .
Proof.
Clearly, Man is in . To prove hardness, we reduce from SSS. Given an instance of SSS, we construct an instance of Man with three candidates . Let and . Set . In what follows we describe the votes in each district as a list . The districts are partitioned into three sets , and :

has a district with votes for each , and a district with votes for each .

consists of a single district with votes .

consists of three districts with votes , , and .
For every district we set . The attacker is allowed to change all votes in each district in and , but none in . Finally, let and . The true winner in this profile is candidate with weight , compared to the weight of and of .
Given a set of integers , let be the corresponding set of districts in . Assume that there is a subset with such that no has sum equal to . The attacker can then exchange the weights of and in the districts in and the district in . This way, becomes the winner with weight , compared to the weight of and the weight of .
Since , to defeat the attacker, the defender needs to restore as the winner. To this end, she must recount the district in , as otherwise ’s weight will remain at least . Hence she can recount at most manipulated districts in . Let the set of nonrecounted districts in be for some ; note that , so by assumption, . Then, the weight of is and the weight of is . At least one of these numbers is greater than or equal to ; thus, cannot be restored as the winner.
Conversely, suppose that for every subset of size there exists a nonempty such that . Then, the attacker cannot win. Indeed, let be the set of manipulated districts. If a district is changed in favor of , the defender can recount all other districts in . On the other hand, if all districts in are won by or , the defender can identify a nonempty subset of such that the corresponding integers sum up to , and request a recount of all other districts in . Such a recount recovers the correct weights of and , and is restored as the winner. ∎
We conjecture that Man remains complete when the input is given in unary; however, for this setting we are only able to prove that this problem is NPhard.
Theorem 4.7.
Man is NPhard, even when and the input vote profile and district weights are given in unary.
Proof.
To show that Man is NPhard even when the input votes and district weights are given in unary, we provide a reduction from Independent Set; see Definition 2.4.
Given an instance of Independent Set with , we construct the following instance of Man. Let , . We set , where is the attacker’s preferred candidate; thus, . Then, we create the following districts; the weight of each district is equal to the number of voters therein.

For every edge , we create two districts and with voters each; thus, . In each such district there are two voters who vote for and three voters who vote for . We set ; thus, the attacker can change the winner in this district from to .

For every node , we create a district with voters; thus, . In each such district there are voters who vote for and voters who vote for . We set ; thus, the attacker can change the winner in this district from to .

There are also some districts that cannot be manipulated (i.e., ). We specify the weights and the winners of these districts.

For each , there is a district with weight and winner .

For each , there is a district with weight and winner .

Finally, there is a district with weight and winner .

The budgets are and .
We have , , for each , and for each . Hence, the true winner of the election is candidate . We show that the attacker can make the winner if and only if is a yesinstance of Independent Set, i.e., there is an independent set of size in .
Suppose first that there is an independent set , , in . The following manipulation strategy makes the winner. For every , change the winner of district from to , and for every such that , change the winner of district from to . Note that since is an independent set, the weight of each candidate , , increases by at most . Let denote the weight of each candidate after manipulation. We have , , for each , and for each ; thus, candidate becomes the winner of the election.
Conversely, suppose that the attacker has a manipulation that makes the election winner; for each , let be the weight of candidate after this manipulation. Since cannot be made the winner in any additional district, we have . Let be the set of all nodes such that the attacker changes the winner of