1 Introduction
We consider a committee voting setting in which each voter approves of a subset of candidates and based on the approvals, a target number of candidates are selected. The setting has been referred to as approvalbased multiwinner voting or committee voting with approvals. The setting has inspired a number of natural voting rules (Kilgour, 2010; Brams and Fishburn, 2007; LeGrand et al., 2007; Aziz et al., 2015b; Skowron et al., 2016). Many of the voting rules attempt to satisfy some notion of representation. However it has been far from clear what axiom captures the representation requirements.
Aziz et al. (2015a, 2017) proposed two compelling representation axioms called justified representation () and extended justified representation (). Whereas the former can be tested as well as achieved in polynomial time, the latter property is coNPcomplete to test and no polynomialtime algorithm is known to achieve it. Interestingly, SánchezFernández et al. (2016a) proposed an intermediate property called proportional justified representation () that admits a polynomialtime algorithm to achieve (Brill et al., 2017; SánchezFernández et al., 2016b).^{1}^{1}1The property was independently proposed by Haris Aziz in October 2014 who referred to it as weak . The idea behind all the three properties is that a cohesive and large enough group deserves sufficient number of approved candidates in the winning set of candidates. SánchezFernández et al. (2016a) argued that although is a stronger property than , is more reasonable because it is compatible with a property called perfect representation.
Proportional justified representation () has been examined in subsequent papers (Brill et al., 2017; SánchezFernández et al., 2016b; SánchezFernández et al., 2017). Despite the flurry of work on the property, the complexity of testing proportional justified representation has remained an open problem. SánchezFernández et al. (2017) state that “we do not know what is the complexity of checking whether a given committee provides PJR”. In a talk “Approval Voting, representation, & Liquid Democracy” at the Workshop on Future Directions in Computational Social Choice, Hungary in November 2016, Markus Brill also mentioned the problem as an interesting open problem. ^{2}^{2}2http://econ.core.hu/file/download//future_markus.pdf The problem is especially important if one wants to test whether a status quo outcome or the outcome of some other rule or negotiation process satisfies . Previously, Aziz et al. (2016) studied the complexity of testing Pareto optimality of a committee.
In this paper, we settle the complexity of testing by proving that the problem is coNPcomplete. We complement the complexity result by showing that the problem admits efficient algorithms if any of the following parameters are bounded: (1) (number of voters) (2) (number of candidates) (3) (maximum number of candidates approved by a voter) (4) (maximum number of voters approving a given candidate). For the first two parameters, we show that the problem is FPT (fixedparameter tractable), i.e, there exists an FPT algorithm that solves the problem in time, where where is the parameter and is some computable function and is a polynomial both independent of problem instance .
Our results are summarized in Table 1.
2 Approvalbased Committee Voting and Representation Properties
We consider a social choice setting with a set of voters and a set of candidates. Each voter submits an approval ballot , which represents the subset of candidates that she approves of. We refer to the list of approval ballots as the ballot profile. We will consider approvalbased multiwinner voting rules that take as input a tuple , where is a positive integer that satisfies , and return a subset of size , which we call the winning set, or committee.
We now summarize the main representation properties proposed in the literature.
Definition 1 (Justified representation (JR)).
Given a ballot profile over a candidate set and a target committee size , we say that a set of candidates of size satisfies justified representation for if
The rationale behind this definition is that if candidates are to be selected, then, intuitively, each group of voters “deserves” a representative. Therefore, a set of voters that have at least one candidate in common should not be completely unrepresented.
Definition 2 (Extended justified representation (EJR)).
Given a ballot profile over a candidate set , a target committee size , , we say that a set of candidates , , satisfies extended justified representation for and integer if
We say that satisfies extended justified representation for if it satisfies satisfies extended justified representation for and all integers .
SánchezFernández et al. (2016a) came up with the notion of proportional justified representation (), which can be seen as an alternative to .
Definition 3 (Proportional Justified Representation ()).
Given a ballot profile over a candidate set , a target committee size , , and integer we say that a set of candidates , , satisfies proportional justified representation for if
We say that satisfies proportional justified representation for if it satisfies satisfies proportional justified representation for and all integers .
It is easy to observe that EJR implies PJR which implies JR.
3 Results
We first prove that testing is coNPcomplete. The proof involves a similar type of reduction as the one used by Aziz et al. (2015a, 2017) to prove that testing is coNPcomplete.
Theorem 1.
Given a ballot profile , a target committee size , and a committee , , it is coNPcomplete to check whether satisfies for .
Proof.
It is easy to see that this problem is in coNP. A set of voters such that , and is a certificate that does not satisfy .
To prove coNPcompleteness, we reduce the classic Balanced Biclique problem ([GT24] in Garey and Johnson 1979) to the complement of our problem. An instance of Balanced Biclique is given by a bipartite graph with parts and and edge set , and an integer ; it is a “yes”instance if we can pick subsets of vertices and so that and for each ; otherwise, it is a “no”instance.
Given an instance of Balanced Biclique with , we create an instance of our problem as follows. Assume without loss of generality that , . We construct pairwise disjoint sets of candidates , and , so that , , , and set . We then construct sets of voters , , , so that , , (note that as we assume that ). For each we set , and for each we set . The candidates in are matched to voters in : each voter in approves exactly one candidate in , and each candidate in is approved by exactly one voter in . Denote the resulting list of ballots by . Finally, we set , and let , where is a subset of with . Note that the number of voters is given by , so .
Suppose first that we started with a “yes”instance of Balanced Biclique, and let be the respective by biclique. Let and . Then , all voters in approve all candidates in , , but all voters in together are only represented by candidates in . Hence, fails to provide proportional justified representation for .
Conversely, suppose that fails to provide for . That is, there exists a value , a set of voters and a set of candidates so that all voters in approve of all candidates in , but all voters in together are only represented by less than candidates in . Note that, since and , we have . Further, since and , it follows that contains one voter from . So, all voters in together are represented by exactly candidates in . This implies that . As , it follows that . Since contains voters from both and , it follows that . Thus, there are at least voters in who approve the same candidates in ; any set of such voters and such candidates corresponds to an by biclique in the input graph. ∎
Note that although there is a polynomialtime algorithm to compute a committee that achieves (Brill et al., 2017; SánchezFernández et al., 2016b), we have proved that checking whether any arbitrary committee achieves is coNPcomplete. We complement the negative computational result by showing that testing is computationally tractable if one of the following parameters is bounded.



(maximum size of approval sets).

(maximum number of approvals of a candidate).
We first observe that testing is in FPT with parameter .
Theorem 2.
Testing is in FPT with parameter and takes time at most .
Proof.
Suppose we want to check whether satisfies . If is bounded then one can simply brute force all the possible violating sets of voters and check that if then it must be that . ∎
Next, we prove that testing is in FPT with parameter .
Theorem 3.
Testing is in FPT with parameter and takes time at most .
Proof.
Suppose we want to check whether satisfies . Note that it is sufficient to show that testing  is in FPT with parameter . Note that for all . Since , .
We go through all the subsets of size . Each set is viewed as the intersection of possible objecting/deviating set of voters. For each , we find the corresponding set of voters as follows:
We return no (i.e., does not satisfy ) if , but . If we do not return no for any , in that case we return yes (i.e., satisfies ).
We now argue that it takes at most operations to test . To check , we go through sets. For each set , we find which takes steps. After that we find which takes an additional operations. Hence it takes operations to test  and it takes operations to test . ∎
If is bounded, then can be tested in polynomial time.
Theorem 4.
If is bounded, testing is solvable in polynomial time .
Proof.
Suppose we want to check whether satisfies . Note that it is sufficient to show that testing  is polynomialtime solvable for each if is bounded. Note that we only need to consider because the maximum size of intersection of any set of approval sets is at most which means that . For larger than ,  is trivially satisfied.
We now describe the algorithm to test  for all . We go through all the subsets of size . There are at most such sets. Since and is bounded, it implies that is constant as well and hence there at most different subsets to be considered.
Each set is viewed as the intersection of possible objecting /deviating set of voters. For each , we find the corresponding set of voters as follows:
We return no (i.e., does not satisfy ) if , but . If we do not return no for any , in that case we return yes (i.e., satisfies ).
We now argue that it takes at most operations to test . To check , we go through at most sets. For each set , we find which takes steps. After that we find which takes an additional operations. Hence it takes operations to test  and it takes operations to test . ∎
Finally, we show that if is bounded, then can be tested in polynomial time.
Theorem 5.
If is bounded, testing is solvable in polynomial time .
Proof.
Suppose we want to check whether satisfies . Note that for any deviating/objecting set of voters , . The reason is that any candidate is approved by at most voters. Hence in order to check for a violation of , we just need to check for sets of voters of size at most . There are at most such set of voters. For each such coalition of voters, we just need to check for  for . So . For each among the at most sets of voters, we need to test for  which requires us to compute , , and . Hence testing  of takes time and testing takes time . ∎
In this paper, we examined the complexity of testing , an interesting new axiom in committee voting. The arguments for all of our positive algorithmic results also hold for testing rather than . It will be interesting to see whether testing is FPT with respect to parameters or .
References

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