Structure in Dichotomous Preferences
SStructure in Dichotomous Preferences ∗ Edith ElkindUniversity of Oxford, [email protected] Martin LacknerUniversity of Oxford, [email protected]
Abstract
Many hard computational social choice problems are known to become tractablewhen voters’ preferences belong to a restricted domain, such as those of single-peaked or single-crossing preferences. However, to date, all algorithmic results ofthis type have been obtained for the setting where each voter’s preference list is atotal order of candidates. The goal of this paper is to extend this line of research tothe setting where voters’ preferences are dichotomous , i.e., each voter approvesa subset of candidates and disapproves the remaining candidates. We proposeseveral analogues of the notions of single-peaked and single-crossing preferencesfor dichotomous profiles and investigate the relationships among them. We thendemonstrate that for some of these notions the respective restricted domains admitefficient algorithms for computationally hard approval-based multi-winner rules.
Preference aggregation is a fundamental problem in social choice, which has recentlyreceived a considerable amount of attention from the AI community. In particular,an important research question in computational social choice (Brandt et al., 2015) isthe complexity of computing the output of various preference aggregation procedures.While for most common single-winner rules winner determination is easy, many at-tractive rules that output a committee (a fixed-size set of winners) or a ranking of thecandidates are known to be computationally hard.There are several ways to circumvent these hardness results, such as using ap-proximate and parameterized algorithms. These standard algorithmic approaches arecomplemented by an active stream of research that analyzes the computational com-plexity of voting rules on restricted preference domains , such as the classic domainsof single-peaked (Black, 1958) or single-crossing (Mirrlees, 1971) preferences. Thisresearch direction was popularized by Walsh (2007) and Faliszewski et al. (2011), andhas lead to a number of efficient algorithms for winner determination under prominentvoting rules as well as for manipulation and control, which can be used when voters’preferences belong to one of these restricted domains (Walsh, 2007; Faliszewski et al., ∗ A preliminary version appeared in the proceedings of IJCAI 2015, the International Joint Conference onArtificial Intelligence (Elkind and Lackner, 2015). a r X i v : . [ c s . G T ] F e b dichotomous , i.e., each voter approves a subset of thecandidates and disapproves the remaining candidates. Committee selection rules forvoters with dichotomous preferences, or approval-based rules, have recently attractedsome attention from the computational social choice community, and for two promi-nent such rules (specifically, Proportional Approval Voting (PAV) (Kilgour and Mar-shall, 2012) and Maximin Approval Voting (MAV) (Brams et al., 2007)) computingthe winning committee is known to be NP-hard (Aziz et al., 2015a; LeGrand et al.,2007). It is therefore natural to ask if one could identify a suitable analogue of single-peaked/single-crossing preferences for the the dichotomous setting, and design effi-cient algorithms for approval-based rules over such restricted dichotomous preferencedomains.To address this challenge, in this paper we propose and explore a number of domainrestrictions for dichotomous preferences that build on the same intuition as the conceptsof single-peakedness and single-crossingness. Some of our restricted domains are de-fined by embedding voters or candidates into the real line, and requiring that the voters’preferences over the candidates “respect” this embedding; others are obtained by view-ing dichotomous preferences as weak orders and requiring them to admit a refinementthat has a desirable structural property. Surprisingly, these approaches lead to a largenumber of concepts that are pairwise non-equivalent and capture different aspects ofour intuition about what it means for preferences to be “one-dimensional”. We analyzethe relationships among these restricted preference domains, (see Figure 5 for a sum-mary), and discuss the complexity of detecting whether a given dichotomous profilebelongs to one of these domains. We then demonstrate that considering these domainsis useful from the perspective of algorithm design, by providing polynomial-time andFPT algorithms for PAV and MAV under some of these domain restrictions. Let C = { c , . . . , c m } be a finite set of candidates. A partial order (cid:31) over C is areflexive, antisymmetric and transitive binary relation on C ; a partial order (cid:31) is said tobe total if for each c, d ∈ C we have c (cid:31) d or d (cid:31) c . We say that a partial order (cid:31) over C is a dichotomous weak order if C can be partitioned into two disjoint sets C + and C − (one of which may be empty) so that c (cid:31) d for each c ∈ C + , d ∈ C − and thecandidates within C + and C − are incomparable under (cid:31) .An approval vote on C is an arbitrary subset of C . We say that an approval vote v is trivial if v = ∅ or v = C . A dichotomous profile P = ( v , . . . , v n ) is a list of n approval votes; we will refer to v i as the vote of voter i . We write v i = C \ v i .We associate an approval vote v i with the dichotomous weak order (cid:31) v i that satisfies c (cid:31) v i d if and only if c ∈ v i , d ∈ v i . Note that v i = ∅ and v i = C correspond to the2ame dichotomous weak order, namely the empty one.A partial order (cid:31) (cid:48) over C is a refinement of a partial order (cid:31) over C if for every c, d ∈ C it holds that c (cid:31) d implies c (cid:31) (cid:48) d . A profile P (cid:48) = ( (cid:31) , . . . , (cid:31) n ) of totalorders is a refinement of a dichotomous profile P = ( v , . . . , v n ) if (cid:31) i is a refinementof (cid:31) v i for each i = 1 , . . . , n .Let (cid:67) be a total order over C . A total order (cid:31) over C is said to be single-peakedwith respect to (cid:67) if for any triple of candidates a, b, c ∈ C with a (cid:67) b (cid:67) c or c (cid:67) b (cid:67) a itholds that a (cid:31) b implies b (cid:31) c . A profile P of total orders over C is said to be single-peaked if there exists a total order (cid:67) over C such that all orders in P are single-peakedwith respect to (cid:67) .A profile P = ( (cid:31) , . . . , (cid:31) n ) of total orders over C is said to be single-crossing withrespect to the given order of votes if for every pair of candidates a, b ∈ C such that a (cid:31) b all votes where a is preferred to b precede all votes where b is preferred to a ; P is single-crossing if the votes in P can be permuted so that it becomes single-crossingwith respect to the resulting order of votes.A profile P = ( (cid:31) , . . . , (cid:31) n ) of total orders over C is said to be -Euclidean ifthere is a mapping ρ of voters and candidates into the real line such that c (cid:31) i d if andonly if | ρ ( i ) − ρ ( c ) | < | ρ ( i ) − ρ ( d ) | . A -Euclidean profile is both single-peaked andsingle-crossing. We will now define a number of constraints that a dichotomous profile may satisfy.Most of these constraints can be divided into two basic groups: those that are basedon ordering voters and/or candidates on the line and requiring the votes to respect thisorder (this includes VEI, VI, CEI, CI, DE, and DUE), and those that are based onviewing votes as weak orders and asking if there is a single-peaked/single-crossing/1-Euclidean profile of total orders that refines the given profile (this includes PSP, PSC,and PE); we remark that the study of the latter type of constraints was initiated byLackner (2014). We will also consider constraints that are based on partitioning vot-ers/candidates (2PART and PART), as well as two constraints (WSC and SSC) thathave been introduced in a recent paper of Elkind et al. (2015) in order to understandthe best way of extending the single-crossing property to weak orders.Fix a profile P = ( v , . . . , v n ) over C .1. -partition (2PART) : We say that P satisfies 2PART if P contains only two dis-tinct votes v, v (cid:48) , and v ∩ v (cid:48) = ∅ , v ∪ v (cid:48) = C .2. Partition (PART) : We say that P satisfies PART if C can be partitioned into pair-wise disjoint subsets C , . . . , C (cid:96) such that { v , . . . , v n } = { C , . . . , C (cid:96) } (i.e.,each voter in P approves one of the sets C , . . . , C (cid:96) ). Note that this constraintcontains as a special case profiles where every voter approves of exactly onecandidate.3. Voter Extremal Interval (VEI) : We say that P satisfies VEI if the voters in P canbe reordered so that for every candidate c the voters that approve c form a prefix3 v v b c v v v a d Figure 1: Voter Extremal Interval v v v v v v a db c Figure 2: Voter Intervalor a suffix of the ordering. Equivalently, both the voters who approve c and thevoters who disapprove c form an interval of that ordering. See Figure 1 for anexample.4. Voter Interval (VI) : We say that P satisfies VI if the voters in P can be reorderedso that for every candidate c the voters that approve c form an interval of thatordering. See Figure 2 for an example. Candidate Extremal Interval (CEI) : We say that P satisfies CEI if candidates in C can be ordered so that each of the sets v i forms a prefix or a suffix of thatordering. Equivalently, both v i and v i form an interval of that ordering. SeeFigure 3 for an example.5. Candidate Interval (CI) : We say that P satisfies CI if candidates in C can beordered so that each of the sets v i forms an interval of that ordering. See Figure 4for an example.6. Dichotomous Uniformly Euclidean (DUE) : We say that P satisfies DUE if thereis a mapping ρ of voters and candidates into the real line and a radius r such thatfor every voter i it holds that v i = { c : | ρ ( i ) − ρ ( c ) | ≤ r } .7. Dichotomous Euclidean (DE) : We say that P satisfies DE if there is a mapping ρ of voters and candidates into the real line such that for every voter i there existsa radius r i with v i = { c : | ρ ( i ) − ρ ( c ) | ≤ r i } .8. Possibly single-peaked (PSP) : We say that P satisfies PSP if there is a single-peaked profile of total orders P (cid:48) that is a refinement of P .9. Possibly single-crossing (PSC) : We say that P satisfies PSC if there is a single-crossing profile of total orders P (cid:48) that is a refinement of P . c c c c c c v v v v Figure 3: Candidate Extremal Interval c c c c c c v v v v Figure 4: Candidate Interval4PART PARTPSC=SSCVEI CEIWSC DUECI=DE=PSP=PEVIFigure 5: Relations between notions of structure. Dashed lines indicate that the respec-tive containment holds only subject to additional conditions.10.
Possibly Euclidean (PE) : We say that P satisfies PE if there is a -Euclideanprofile of total orders P (cid:48) that is a refinement of P .11. Seemingly single-crossing (SSC) : We say that P satisfies SSC if the voters in P can be reordered so that for each pair of candidates a, b ∈ C it holds that eitherall votes v i with a ∈ v i , b (cid:54)∈ v i precede all votes v j with a (cid:54)∈ v j , b ∈ v j or viceversa.12. Weakly single-crossing (WSC) : We say that P satisfies WSC if the voters in P can be reordered so that for each pair of candidates a, b ∈ C it holds that eachof the vote sets V = { v i : a ∈ v i , b (cid:54)∈ v i } , V = { v i : a (cid:54)∈ v i , b ∈ v i } , V = { v ∈ P : v (cid:54)∈ V ∪ V } forms an interval of this ordering, with V appearing between V and V . The relationships among the properties defined above are depicted in Figure 5, wherearrows indicate containment, i.e., more restrictive notions are at the top. All thesecontainments are strict.The four arrows at the top level of the diagram are immediate: any profile with atmost two distinct votes where each candidate is approved in at least one of these votessatisfies VEI, CEI and WSC, and by definition 2PART is a special case of PART.To understand the arrows in the next level, we first characterize the dichotomousprofiles that are weakly single-crossing.
Lemma 1.
A dichotomous profile P satisfies WSC if and only if there exist three votes u, v, w such that(1) for every v i ∈ P it holds that (cid:31) v i ∈ {(cid:31) u , (cid:31) v , (cid:31) w } , and(2) (cid:31) v is equal to either (cid:31) u ∩ w or (cid:31) u ∪ w . roof sketch. It is easy to check that every profile satisfying (1)–(2) satisfies WSC. Forthe converse direction, assume without loss of generality that the ordering of the votes v (cid:64) v (cid:64) · · · (cid:64) v n witnesses that P satisfies WSC. Let u = v , w = v n , and set C = u ∩ w , C = u ∩ w , C = u ∩ w , C = u ∩ w . The WSC property impliesthat for every (cid:96) = 1 , , , , every a, b ∈ C (cid:96) , and every v i ∈ P we have a ∈ v i ifand only if b ∈ v i , i.e., candidates in each C (cid:96) occur as a block in all votes. Note that v = u = C ∪ C , v n = w = C ∪ C .Suppose that C , C (cid:54) = ∅ . Then C ⊆ v i , C ⊆ v i for all v i ∈ P . Indeed, fix a pairof candidates a ∈ C , b ∈ C . Both the first and the last voter strictly prefer a to b , andtherefore so do all other voters. Thus, if P contains a vote v i (cid:54) = u, w , it has to be thecase that v i = C = u ∩ w or v i = C ∪ C ∪ C = u ∪ w ; moreover, if both of thesevotes occur simultaneously and are distinct from each other and u, w (i.e., C , C (cid:54) = ∅ ),the WSC property is violated. Indeed, suppose that v i = C , v j = C ∪ C ∪ C . Fixcandidates a ∈ C , b ∈ C . If v i appears before v j , consider a candidate c ∈ C : weget a contradiction as voters v and v j are indifferent between a and c , but v i strictlyprefers a to c . If v i appears after v j , consider a candidate d ∈ C : we get a contradictionas voters v and v i are indifferent between d and b , but v j strictly prefers d to b . When C or C is empty, the analysis is similar; note, however, that trivial votes ( v i = C and v i = ∅ ) may alternate arbitrarily without violating the WSC property (this is why thelemma is stated in terms of weak orders rather than approval votes).We can now show that under mild additional conditions (no trivial voters/candidates)WSC implies VEI and CEI. Proposition 2.
Let P be a dichotomous profile that either contains only two distinctvotes or contains no vote v i with v i = ∅ . If P satisfies WSC, then it satisfies VEI.Proof. Assume without loss of generality that P satisfies WSC with respect to an or-dering of voters v (cid:64) · · · (cid:64) v n , and let u = v , w = v n . We will show that P satisfiesVEI with respect to (cid:64) . If P only contains two distinct votes, this claim is immediate,so assume that ∅ (cid:54)∈ P . Consider a vote v ∈ P that is distinct from u and w . Since ∅ (cid:54)∈ P , by Lemma 1 there exist i, j with < i < j < n such that v k = u for k < i , v k = v for k = i, . . . , j , v k = w for k > j , and v ∈ { u ∪ w, u ∩ w } . Suppose first that v = u ∩ w . Then candidates in u ∩ w are approved by all voters, candidates in u \ w are approved by the first i − voters, candidates in w \ u are approved by the last n − j voters, and the remaining candidates are not approved by anyone. On the other hand,if v = u ∪ w , then candidates in u ∩ w are approved by all voters, candidates in u \ w are approved by the first j voters, candidates in w \ u are approved by the last n − i + 1 voters, and the remaining candidates are not approved by anyone.The condition that the profile must not contain ∅ is necessary: the profile ( { a, b } , ∅ , { b, c } ) satisfies WSC, but not VEI. Proposition 3.
Let P be a dichotomous profile that either contains only two distinctvotes or in which every candidate is approved in at least one vote and disapproved inat least one vote. If P satisfies WSC, then it satisfies CEI. roof. Suppose that P is WSC with respect to some ordering of voters; let u and w be, respectively, the first and the last vote in this ordering. If P contains a trivial vote,it contains at most two non-trivial votes, in which case the claim is obvious. Thus,assume that it contains no trivial votes. Then we have u ∩ w = ∅ (any candidate in u ∩ w would be approved by all voters) and u ∩ w = ∅ (any candidate in u ∩ w wouldbe disapproved by all voters). It is now easy to see that ordering the candidates so thatall candidates approved by u precede all candidates approved by w witnesses that P isCEI.To see that conditions of Proposition 3 are necessary, consider the profile ( { a, b } , { b, c } ) over { a, b, c, d } and the profile ( { a, b } , { b } , { b, c } ) over { a, b, c } : both of these profilessatisfy WSC, but not CEI.Interestingly, requiring a dichotomous profile to satisy WSC, CEI and VEI simul-taneously, turns out to be very demanding: we obtain 2-partition profiles. Proposition 4.
A dichotomous profile is WSC, CEI and VEI if and only if it is a -partition.Proof. It is immediate that a -partition profile is WSC, CEI, and VEI. For the conversedirection, let P be a CEI, VEI and WSC profile. By Lemma 1, P contains at mostthree distinct votes u, v, w with v = u ∩ w or v = u ∪ w . Since P is CEI, we knowfrom Lemma 3 that every candidate is approved at least once. Hence u ∪ w = C .Furthermore, every candidate is disapproved at least once. Thus, u ∩ w = ∅ , since thisintersection is also approved by v . Thus, v is a trivial vote. This is possible becauseof Lemma 2 and hence v does not appear in P . We have shown that P is a -partitionprofile.Next, we will relate CEI and VEI to DUE. Proposition 5.
If a dichotomous profile P satisfies CEI or VEI, then it satisfies DUE.Proof. Suppose first that P satisfies CEI with respect to the ordering c (cid:67) · · · (cid:67) c m of candidates. Map the candidates into the real line by setting ρ ( c i ) = i , and let r = m . We can now place each voter i to the left or to the right of all candidates at anappropriate distance so that the set of candidates within distance r from him coincideswith v i . For VEI the argument is similar: if P satisfies VEI with respect to the ordering v (cid:64) · · · (cid:64) v n of voters, we place voters on the real line according to ρ ( i ) = i , let r = n , and place each candidate to the left or to the right of all voters at an appropriatedistance.The proof that WSC implies DUE is also based on our characterization of WSCpreferences. Proposition 6.
If a dichotomous profile P satisfies WSC, then it satisfies DUE.Proof. Clearly empty votes can be ignored when checking whether a profile satisfiesDUE, so assume P contains to empty votes. Then it contains at most three distinctvotes u , v , w with v = u ∩ w or v = u ∪ w . Set ρ ( c ) = 1 for c ∈ u \ w , ρ ( c ) = 2 for c ∈ u ∩ w , ρ ( c ) = 3 for c ∈ w \ u , ρ ( c ) = 10 for c (cid:54)∈ u ∪ w . We set r = 1 if v = u ∩ w and r = 2 if v = u ∪ w , and position the voters accordingly.7he last arrow on this level is from PART to DUE: here, the containment is straight-forward, as the candidates approved by each voter can be placed as a block on the axis,with the respective voter(s) placed in the center of this block. Proposition 7.
If a dichotomous profile P satisfies DUE then it satisfies both VI andCI. The converse direction does not hold: there are profiles that satisfy VI and CI butnot DUE.Proof. Since P satisfies DUE, we have an embedding ρ of votes and candidates intothe real line. For VI, we order voters as induced by the ρ mapping; the voters approvingsome candidate form an interval on this induced order. For CI, we order candidates asinduced by the ρ mapping; voters always approve a single interval on this ordering.For showing that the converse direction does not hold, consider the profile ( { a, b, c } , { b, c, d } , { b } , { c } ) . Towards a contradiction assume that ρ is a mapping of voters andcandidates into the real line that witnesses the DUE property for a fixed radius r . Thegiven profile satisfies CI only with respect to the orders a (cid:67) b (cid:67) c (cid:67) d , a (cid:67) c (cid:67) b (cid:67) d andtheir reverses. Since the profile is symmetric with respect to a and d and with respect to b and c , we can assume without loss of generality that ρ orders candidates as the order a (cid:67) b (cid:67) c (cid:67) d does. Then it has to hold that | ρ ( a ) − ρ ( c ) | ≤ r since a and c appear inthe same vote. However, due to the vote { b } , it also has to hold that | ρ ( a ) − ρ ( c ) | > r ;this is a contradiction.We see that similar to total orders, where the intersection of the single-peaked andthe single-crossing domain is a strict subset of the 1-Euclidean domain (see discussionin (Doignon and Falmagne, 1994; Elkind et al., 2014)), for dichotomous preferencesalso VI intersected with CI does not yield DUE. The next results shows that the classesof CI, DE, PSP and PE preferences coincide. Proposition 8.
Let P be a dichotomous profile. Then the following conditions areequivalent: (a) P satisfies PE (b) P satisfies PSP (c) P satisfies CI (d) P satisfies DE.Proof sketch. Suppose P satisfies PE, and let P (cid:48) be a refinement of P that, togetherwith a mapping ρ , witnesses this. Then P (cid:48) is single-peaked and therefore P satisfiesPSP. If P satisfies PSP, as witnessed by a refinement P (cid:48) and an axis (cid:67) , then P satisfiesCI with respect to (cid:67) . If P satisfies CI with respect to an order (cid:67) of candidates, wecan map the candidates into the real axis in the order suggested by (cid:67) so that the dis-tance between every two adjacent candidates is . We can then choose an appropriateapproval radius and position for each voter. Finally, if P satisfies DE, as witnessed bya mapping ρ , we can use this mapping to construct a refinement of P ; by construction,this refinement is 1-Euclidean (we may have to modify ρ slightly to avoid ties).Also, every PE profile is PSC since every 1-Euclidean refinement is also single-crossing. Interestingly, the converse is not true. Example 1.
Consider the profile P = ( { a, b } , { a, c } , { b, c } ) over C = { a, b, c } . Itsatisfies PSC, as witnessed by the single-crossing refinement ( a (cid:31) b (cid:31) c, c (cid:31) a (cid:31) b, c (cid:31) b (cid:31) a ) . However, in every refinement of P the first voter ranks c last, the secondvoter ranks b last, and the third voter ranks a last. Thus, no such refinement can besingle-peaked, and, consequently, no such refinement can be 1-Euclidean.8he equivalence between PSC and SSC is not entirely obvious: while it is clear thata profile that violates SSC also violates PSC, to prove the converse one needs to usean argument similar to the proof of Theorem 4 in (Elkind et al., 2015). This has beenshown in the extended version of (Elkind et al., 2015). Proposition 9.
If a dichotomous profile P satisfies VI, it also satisfies SSC.Proof. Assume that an VI profile is not SSC. Since it is not SSC, for every ordering ofvotes (cid:64) there are two candidates a (cid:31) b and votes v i (cid:64) v j (cid:64) v k such that v i : a (cid:31) b , v j : b (cid:31) a and v k : a (cid:31) b . This implies, however, that for every (cid:64) there is a candidate a and votes v i (cid:64) v j (cid:64) v k such that v i and v k approve of a and v j disapproves v j . Thiscontradicts our assumption that the given profile is VI.We are now going to list the remaining counter-examples for containment and thusshow that the arrows in Figure 5 indeed indicate strict containment. • CI (cid:54)→ VI: Consider ( { a, b, c } , { a } , { b } , { c } ) . This profile is CI with respect to a (cid:67) b (cid:67) c . It is not VI since the vote { abc } would have to be placed next to { a } , { b } , { c } . • VI (cid:54)→ CI: Consider ( { a, b } , { a, c } , { a, d } ) . This profile is VI for the given orderof voters. It is not CI since a has to lie next to b, c, d . • VEI (cid:54)→
CEI: Consider ( { a, b } , { a, d } , { c, d } ) . This profile is VEI for the givenorder of voters. It is not CEI since a has to lie next to b and d and c has to lienext to d . So b (cid:67) a (cid:67) d (cid:67) c is the only order witnessing CI, but the vote { a, d } isnot an extremal interval on this order. • CEI (cid:54)→
VEI: Consider ( { a, b } , { a } , { c } , { b, c } ) . All votes are extremal intervalson the order a (cid:67) b (cid:67) c . The profile is however not VEI since { a, b } has to lienext to { a } and next to { bc } and { c } next to { bc } . So we obtain { c } (cid:64) { b, c } (cid:64) { a, b } (cid:64) { a } as the only order witnessing VI, but this order does not satisfy VEI(consider candidate b ). • PART (cid:54)→
VEI, CEI, WSC: Consider the PART profile { a } , { b } , { c } .All other counterexamples involving WSC immediately follow from Lemma 1 andProposition 2 and 3; all missing counterexamples involving PART can be obtained bypicking intersecting votes. If voter’s preferences are given by total orders, single-crossing profiles have a uniquesingle-crossing order, i.e., only one specific order and its reverse witness the single-crossing property of the profile. For single-peaked profiles (of total orders) this isnot the case. The question arises whether a similar phenomenon can be observed fordichotomous profiles. Clearly, this question only makes sense for profiles with distinctvotes (for VI, VEI, WSC) and when all candidates are approved by some vote (for CI9nd CEI). Also, by unique we always mean that only one specific order and its reversewitness a certain restriction.For dichotomous profiles satisfying SC, there is no unique order. The profile ( { a } , { a, b } , { b, c } ) is SC and all votes that put { b, c } at an outermost position witnessthe SC property. Also profiles satisfying VI or CI do not have unique orders witnessingthese properties; e.g., consider {} , { a } , { b } and { a } , { b } , { c } , respectively.For profiles being WSC, VEI or CEI we can show that their corresponding ordersare indeed unique. For profiles satisfying WSC, this follows from Lemma 1; for profilessatisfying either VEI or CEI the uniqueness can be shown as follows. Lemma 10.
For profiles containing distinct votes, VEI orders are unique.Proof.
Without loss of generality assume that (cid:64) · · · (cid:64) n be a VEI order. Assumetowards a contradiction that (cid:64) (cid:48) is another VEI order that is neither (cid:64) nor its reverse.Consequently, there exist three votes v i , v j , v k , i < j < k for which (cid:64) and (cid:64) (cid:48) disagreeon their order in the sense that v j is not in between v i and v k with respect to (cid:64) (cid:48) . Withoutloss of generality let us assume j (cid:64) (cid:48) i (cid:64) (cid:48) k . Let us consider C X for every X ⊆{ i, j, k } being defined as the set of all candidates approved by the votes correspondingto X but not approved by those corresponding to { i, j, k } \ X . For example, C ik arethose candidates approved by c i and c k but not by c j . Since we have a CEI profile and i (cid:64) j (cid:64) k , we know that C ik = C j = ∅ . Under our assumption that (cid:64) (cid:48) is also aVEI ordering with j (cid:64) (cid:48) i (cid:64) (cid:48) k , we know that C jk = C i = ∅ . This implies that thecandidate approved by c i are C ijk ∪ C ij and those approved by c j are also C ijk ∪ C ij .This contradicts our assumption that all votes are distinct. Lemma 11.
If all candidates are approved by distinct sets of voters, CEI orders areunique.Proof.
First, let us observe that two candidates that are approved by the same voterscertainly are indistinguishable; their positions on the CEI axis are interchangeable.Thus, our condition is necessary for the lemma to hold. The proof of this statementis similar to the previous proof. Without loss of generality assume that c (cid:67) · · · (cid:67) c m be a CEI order. Assume towards a contradiction that (cid:67) (cid:48) is another CEI order that isneither (cid:67) nor its reverse. Consequently, there exist three votes v i , v j , v k , i < j < k for which (cid:67) and (cid:67) (cid:48) disagree on their order in the sense that c j is not in between c i and c k with respect to (cid:67) (cid:48) . Without loss of generality let us assume c j (cid:67) (cid:48) c i (cid:67) (cid:48) c k .Let us consider V X for every X ⊆ { i, j, k } being defined as the set of all votes thatapprove the candidates in X and disapprove those in { i, j, k } \ X . Since we have aVEI profile and c i (cid:67) c j (cid:67) c k , we know that V ik = V j = ∅ . Under our assumption that (cid:67) (cid:48) is also a CEI ordering with c j (cid:67) (cid:48) c i (cid:67) (cid:48) c k , we know that V jk = V i = ∅ . This impliesthat the votes that approve c i are V ijk ∪ V ij and the votes approving c j are V ijk ∪ V ij .This contradicts our assumption that all candidates are approved by a distinct set ofvoters. To exploit the constraints defined in Section 3, we have developed algorithms that candecide whether a given profile belongs to one of the restricted domains defined by these10onstraint complexity2PART poly (trivial)PART poly (trivial)VEI poly (
CONSECUTIVE S )CEI poly ( CONSECUTIVE S )WSC poly (Elkind et al., 2015)DUE poly (Nederlof and Woeginger, 2015)VI poly ( CONSECUTIVE S )CI=DE=PSP=PE poly ( CONSECUTIVE S )PSC=SSC openTable 1: The complexity of detecting structure in dichotomous profilesconstraints. Our results are summarized in Table 1.Clearly, verifying whether a given profile satisfies 2PART or PART is straightfor-ward. For most of the remaining problems, we can proceed by a reduction to theclassic C ONSECUTIVE S problem (Booth and Lueker, 1976). This problem asks ifthe columns of a given - matrix can be permuted in such a way that in each row ofthe resulting matrix the s are consecutive, i.e., the s form an interval in each row; itadmits a linear-time algorithm (Booth and Lueker, 1976). Theorem 12.
Detecting whether a dichotomous profile satisfies CEI, CI, VI or VEI ispossible in O ( m · n ) time.Proof. Let C = { c , c , . . . , c m } and P = ( v , v , . . . , v n ) . We construct an instanceof C ONSECUTIVE S in slightly different ways, depending on the property we want todetect. In all cases, we obtain a “yes”-instance if and only if the given profile has thedesired property.Let us start with CI. For each vote, we create one row of the matrix: for each i ∈ [ n ] and j ∈ [ m ] , the j -th entry of the i -th row is if c j ∈ v i and otherwise. In this way,we obtain an m × n matrix. Permuting the columns of this matrix so that s form aninterval in each row is equivalent to permuting candidates so that the set of candidatesapproved by each voter forms an interval. For CEI, we combine the matrix for CI withits complement, i.e., we add a second row for each vote v i , so that the j -the entryof that row is of c j ∈ v i and otherwise. A column permutation of the resulting m × n matrix such that s form an interval in each row corresponds to permutingcandidates so that for each voter both the set of her approved candidates and the set ofher disapproved candidates form an interval; this is equivalent to the CEI property. ForVI it suffices to transpose the matrix constructed for CI, and for VEI this matrix has tobe combined with its complement.For WSC, Elkind et al. (2015) provide an algorithm that works for any weak orders(not just dichotomous ones). They leave the complexity of detecting PSC and SSC asan open problem, and we have not been able to resolve it for dichotomous weak orders.The problem of recognizing DUE preferences has recently been shown to be solvable11n polynomial time by Nederlof and Woeginger (2015) via a connection to bipartitepermutation graphs. In this section, we consider two classic approval-based committee selection rules—Proportional Approval Voting (
PAV ) and Maximin Approval Voting (
MAV )—and ar-gue that we can design efficient algorithms for these rules when voters’ preferencesbelong to some of the domains in our list (for some of the richer domains, we mayneed to place mild additional restrictions on voters’ preferences).We start by providing formal definitions of these rules.
Definition 1.
Every non-increasing infinite sequence of non-negative reals w = ( w , w , . . . ) that satisfies w = 1 defines a committee selection rule w - PAV . This ruletakes a set of candidates C , a dichotomous profile P = ( v , . . . , v n ) and a targetcommittee size k ≤ | C | as its input. For every size- k subset W of C , it computes its w - PAV score as (cid:80) v i ∈P u w ( | W ∩ v i | ) , where u w ( p ) = (cid:80) pj =1 w j , and outputs a size- k subset with the highest w - PAV score, breaking ties arbitrarily. The w - PAV rulewith w = (1 , , , . . . ) is usually referred to simply as the PAV rule, and we write u ( p ) = 1 + · · · + p .PAV is of particular interest since it is the only known approval-based committee se-lection rule that satisfies the Extendend Justified Representation property (Aziz et al.,2015b), which intuitively states that every large enough homogenous group has to berepresented in the committee. In what follows we assume that the entries of w arerational and w i can be computed in time poly( i ) . Definition 2.
Given a set of candidates C , a dichotomous profile P = ( v , . . . , v n ) and a target committee size k ≤ | C | , the MAV -score of a size- k subset W of C iscomputed as max v i ∈P ( | W \ v i | + | v i \ W | ) . MAV outputs a size- k subset with thelowest MAV score, breaking ties arbitrarily.The w - PAV rule is defined by Kilgour and Marshall (2012), see also (Kilgour,2010). Intuitively, under this rule each voter is assumed to derive a utility of fromhaving exactly one of his approved candidates in the winning set; his marginal utilityfrom having more of his approved candidates in the winning set is non-increasing. Thegoal of the rule is to maximize the sum of players’ utilities. In contrast, MAV (Bramset al., 2007) has an egalitarian objective: for each candidate committee, it computesthe dissatisfaction of the least happy voter, and outputs a committee that minimizes thequantity.Computing the winning committee under
MAV and
PAV is NP-hard, see, respec-tively, (LeGrand et al., 2007) and (Skowron et al., 2015a; Aziz et al., 2015a). Thehardness result for
PAV extends to w - PAV as long as w satisfies w > w ; more-over, it holds even if each voter approves of at most two candidates or if each candidateis approved by at most three voters.We will now show that PAV admits an algorithm whose running time is polynomialin the number of voters and the number of candidates if the input profile satisfies CI12r VI and, furthermore, each voter approves at most s candidates or each candidate isapproved by at most d voters, where s and d are given constants. More specifically,we prove that PAV winner determination for CI and VI preferences is in FPT withrespect to parameter s and in XP with respect to parameter d . For simplicity, we stateour results for PAV ; however, all of them can be extended to w - PAV .In what follows, we write [ x : y ] to denote the set { z ∈ Z : x ≤ z ≤ y } . Theorem 13.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if | v i | ≤ s for all v i ∈ P and P satisfies VI, then we can find a winning committee under PAV in time O (2 s · k · n ) .Proof. Assume that P satisfies VI with respect to the order of voters v (cid:64) · · · (cid:64) v n .For each triple ( i, A, (cid:96) ) , where i ∈ [1 : n ] , A ⊆ v i , and (cid:96) ∈ [0 : k ] , let r ( i, A, (cid:96) ) bethe maximum utility that the first i voters can obtain from a committee W such that W ∩ v i = A , | W | = (cid:96) , and W ⊆ v ∪ . . . ∪ v i .We have r (1 , A, | A | ) = u ( | A | ) for every A ⊆ v and r (1 , A, (cid:96) ) = −∞ for every A ⊆ v , (cid:96) ∈ [0 : k ] \ {| A |} . To compute r ( i + 1 , A, (cid:96) ) for i ∈ [1 : n − , A ⊆ v i +1 and (cid:96) ∈ [0 : k ] , we let p = | A \ v i | and set r ( i + 1 , A, (cid:96) ) = max D ⊆ v i \ v i +1 r ( i, D ∪ ( A ∩ v i ) , (cid:96) − p ) + u ( | A | ) . Indeed, every committee W with | W | = (cid:96) , W ∩ v i +1 = A , W ⊆ v ∪ . . . ∪ v i +1 contains exactly (cid:96) − p candidates from v ∪ . . . ∪ v i and its intersection with v i is ofthe form D ∪ ( A ∩ v i ) , where candidates in D are approved by v i , but not v i +1 . Weoutput max A ⊆ v n r ( n, A, k ) .This dynamic program has n · s · ( k + 1) states, and the value of each state iscomputed using O (2 s ) arithmetic operations. Assuming that basic calculations takeconstant time, we obtain a total runtime of O (2 s · k · n ) .A similar dynamic programming algorithm can be used if voters’ preferences sat-isfy CI. Theorem 14.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if | v i | ≤ s for all v i ∈ P and P satisfies CI, then we can find a winning committee under PAV in time O (2 s · n · m ) .Proof. Assume that P satisfies CI with respect to the order of candidates c (cid:67) · · · (cid:67) c m .For each triple ( j, A, (cid:96) ) , where j ∈ [1 : m ] , A ⊆ { c j − s +1 , . . . , c j } , and (cid:96) ∈ [0 : k ] , let r ( j, A, (cid:96) ) be the maximum utility that voters can obtain from a committee W such that W ⊆ { c , . . . , c j } , W ∩ { c j − s +1 , . . . , c j } = A , and | W | = (cid:96) . Also, for each j ∈ [1 : m − s + 1] and each A ⊆ { c j , . . . , c j + s − } let t ( A, c j + s − ) = (cid:80) v ∈P : c j + s − ∈ v u ( | A ∩ v | ) . Note that all the quantities t ( ., . ) can be computed in time O (2 s · m · n ) .We have r (1 , ∅ ,
0) = 0 , r (1 , { c } ,
1) = |{ v i : c ∈ v i }| , and r (1 , A, (cid:96) ) = −∞ if ( A, (cid:96) ) (cid:54) = ( ∅ , , ( { c } , . The quantities r ( j + 1 , A, (cid:96) ) for j ∈ [1 : m − can now becomputed as follows. If c j +1 (cid:54)∈ A , we set r ( j + 1 , A, (cid:96) ) = max { r ( j, A, (cid:96) ) , r ( j, A ∪ { c j − s } , (cid:96) ) } . c j +1 ∈ A . Let A (cid:48) = A \{ c j +1 } . Then r ( j +1 , A, (cid:96) ) = max { r , r } where r = r ( j, A (cid:48) ∪ { c j − s } , (cid:96) − − t ( A (cid:48) , c j +1 ) + t ( A, c j +1 ) ,r = r ( j, A (cid:48) , (cid:96) − − t ( A (cid:48) , c j +1 ) + t ( A, c j +1 ) . We output max A ⊆{ c m − s +1 ,...,c m } r ( m, A, k ) . Our dynamic program has at most s · m · ( k + 1) states, and the utility of each state can be computed in time O (1) . Combiningthis with the time used to compute t ( ., . ) , we obtain the desired bound on the runningtime.Our next two theorems also considers CI and VI preferences, and deal with thecase where no candidate is approved by too many voters. Just as the algorithms inthe proofs of Theorems 13 and 14, the algorithms for this case are based on dynamicprogramming. Theorem 15.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if |{ i | c ∈ v i }| ≤ d for all c ∈ C and P satisfies CI, then we can find a winning committee under PAV in time poly( d, m, n, k d ) .Proof. Assume without loss of generality that the candidate order c (cid:67) · · · (cid:67) c m wit-nesses that P is CI. For each voter v i ∈ P , let c b i and c e i be, respectively, the first andthe last candidate (with respect to (cid:67) ) approved by v i , i.e., v i = { c j | b i ≤ j ≤ e i } .For j ∈ [1 : m ] , we say that a voter v i is active at j if b i ≤ j ≤ e i ; we say that a voter v i is finished at j if e i ≤ j . Let B j = { v i | b i = j } , E j = { v i | e i = j } . Given a set W ⊆ C , we will refer to the quantity u ( | W ∩ v i | ) = 1 + 1 / · · · + 1 / | W ∩ v i | as the utility of voter i from set W . Throughout the proof, we make the standard assumptionthat for any real-valued function f we have max { f ( x ) | x ∈ X } = −∞ when X = ∅ .Let R ( j ) be the set of all vectors r ∈ [0 : k ] n such that for all (cid:96) ∈ [1 : n ] it holdsthat ≤ r (cid:96) ≤ min { j − b (cid:96) + 1 , k } and, moreover, r (cid:96) = 0 whenever v (cid:96) is not activeat c j . Vectors in R ( j ) can be used to describe the impact of a set of candidates C in { c , . . . , c j } with | C | ≤ k on voters who are active at c j : for each v (cid:96) ∈ P , r (cid:96) indicateshow many candidates in C are approved by v (cid:96) . As there are at most d voters who areactive at j , we have | R ( j ) | ≤ ( k + 1) d . For each j ∈ [1 : m ] , i ∈ [0 : min { j, k } ] and r ∈ R ( j ) , let W ( i, j, r ) be the collection of all subsets of C with the followingproperties: each W ∈ W ( i, j, r ) satisfies | W | = i , W ⊆ { c , . . . , c j } , and, moreover,for each (cid:96) ∈ [1 : n ] such that v (cid:96) is active at c j it holds that | v (cid:96) ∩ W | = r (cid:96) . Intuitively, W ( i, j, r ) consists of all size- i subsets of { c , . . . , c j } whose impact on voters who areactive at c j is described by r . Let A ( i, j, r ) be the maximum total utility that voterswho are finished at j derive from a set in W ( i, j, r ) ; note that A ( i, j, r ) = −∞ if W ( i, j, r ) = ∅ . Clearly, it is easy to compute A ( i, , r ) for i ∈ { , } and all r ∈ R (1) .We will now explain how to compute A ( i, j, r ) given the values of A ( i (cid:48) , j − , r (cid:48) ) for all i (cid:48) ∈ [0 : min { j − , k } ] and all r ∈ R ( j − .Suppose first that B j (cid:54) = ∅ . By definition of R ( j ) we have r x ∈ { , } for each v x ∈ B j . Moreover, if we have r x (cid:54) = r y for some v x , v y ∈ B j , then W ( i, j, r ) = ∅ and consequently A ( i, j, r ) = −∞ : no subset of { c , . . . , c j } can intersect v x , but not v y or vice versa. 14ow, if B j (cid:54) = ∅ and r x = 1 for all v x ∈ B j , all sets in W ( i, j, r ) contain c j , andtherefore A ( i, j, r ) = max r (cid:48) ∈ R (cid:48) A ( i − , j − , r (cid:48) ) + (cid:88) v (cid:96) ∈ E j u ( r (cid:96) ) , where R (cid:48) is the set of all vectors r (cid:48) ∈ R ( j − with r (cid:48) (cid:96) = r (cid:48) (cid:96) − for all voters v (cid:96) that areactive at both c j and c j − . Indeed, the second summand here is the total utility of votersin E j ; for every such voter v (cid:96) we know that for any set of candidates W ∈ W ( i, j, r ) heapproves exactly r (cid:96) candidates in W . The first summand is the maximum total utilityof voters who are finished at j − that can be achieved by picking a set W (cid:48) so that W (cid:48) ∪ { c j } ∈ W ( i, j, r ) ; every such set W (cid:48) is contained in W ( i − , j − , r (cid:48) ) for somevector r (cid:48) in R ( j − that is consistent with r , i.e. satisfies r (cid:48) (cid:96) = r (cid:48) (cid:96) − for all voters v (cid:96) that are active at both c j and c j − .By a similar argument, if B j (cid:54) = ∅ and r x = 0 for all v x ∈ B j , no set in W ( i, j, r ) contains c j , and therefore A ( i, j, r ) = max r (cid:48) ∈ R (cid:48) A ( i, j − , r (cid:48) ) + (cid:88) v (cid:96) ∈ E j u ( r (cid:96) ) , where R (cid:48) is the set of all vectors r (cid:48) ∈ R ( j − with r (cid:48) (cid:96) = r (cid:48) (cid:96) for all voters v (cid:96) that areactive at both c j and c j − .Finally, suppose that B j = ∅ . Then we have to consider both possibilities for c j .To this end, define a = max r (cid:48) ∈ R (cid:48) A ( i − , j − , r (cid:48) ) + (cid:88) v (cid:96) ∈ E j u ( r (cid:96) ) ,a = max r (cid:48) ∈ R (cid:48) A ( i, j − , r (cid:48) ) + (cid:88) v (cid:96) ∈ E j u ( r (cid:96) ) , where R (cid:48) and R (cid:48) are defined as above, and set A ( i, j, r ) = max { a , a } , again, the argument for correctness is the same as above.To complete the proof, it remains to observe that the PAV -score of an optimalsize- k committee is given by max r ∈ R ( m ) A ( k, m, r ) . Once this score is computed, therespective committee can be found using standard dynamic programming techniques.To bound the running time, note that our dynamic program has O ( km ( k +1) d ) vari-ables, and the argument above establishes that the value of A ( i, j, r ) can be computedin time O ( d ( k + 1) d ) given the values of A ( i (cid:48) , j − , r (cid:48) ) for all i (cid:48) ∈ [0 : min { k, j − } ] , r ∈ R ( j − . Theorem 16.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if |{ i | c ∈ v i }| ≤ d for all c ∈ C and P satisfies VI, then we can find a winning committee under PAV in time poly( d, m, n, k d ) . roof. Assume without loss of generality that the voter order v (cid:64) · · · (cid:64) v n witnessesthat ( C, V ) is in VI. For each candidate c j ∈ C , let v b j and v e j be, respectively, thefirst and the last voter (with respect to (cid:64) ) who approve c j , i.e., { v i ∈ V | c j ∈ v i } = { v i | b j ≤ i ≤ e j } . Let C (cid:96),r = { c j | b j = (cid:96), e j = r } , B (cid:96) = { c j | b j ≤ (cid:96) } . Given a set W ⊆ C , we will refer to the quantity u ( | W ∩ v i | ) = 1 + 1 / · · · + 1 / | W ∩ v i | as the utility of voter i from set W . Throughout the proof, we make the standard assumptionthat for any real-valued function f we have max { f ( x ) | x ∈ X } = −∞ when X = ∅ .Let N ( i ) be the set of all m -by- m matrices over [0 : k ] that have the followingproperty: for every matrix N = ( N (cid:96),r ) (cid:96),r ∈ [1: m ] ∈ N ( i ) , we have ≤ N (cid:96),r ≤ | C (cid:96),r | if (cid:96) ≤ i ≤ r and N r,(cid:96) = 0 if i < (cid:96) or i > r . Matrices in N ( i ) can be used to describe theimpact of a set of candidates W on voter v i : for each (cid:96), r ∈ [1 : m ] , N (cid:96),r indicates howmany candidates in W are approved by v i . Since c j ∈ v i implies i − d + 1 ≤ b j ≤ i , i ≤ e j ≤ i + d − , we have |N ( i ) | ≤ ( k + 1) d .For each j ∈ [0 : k ] , i ∈ [1 : n ] and N ∈ N ( i ) , let W ( i, j, N ) be the collectionof all size- j subsets of B i such that v i ∩ C (cid:96),r = N (cid:96),r for all (cid:96), r ∈ [1 : m ] ; we set W ( i, j, N ) = ∅ if j (cid:54)∈ [0 : k ] , i (cid:54)∈ [1 : n ] or N (cid:54)∈ N ( i ) . In words, W ( i, j, N ) consistsof all size- j sets consisting of candidates that are approved by at least one voter in v , . . . , v i whose impact on v i is described by N . Let A ( i, j, N ) be the maximum totalutility that voters in { v , . . . , v i } derive from a set in W ( i, j, N ) ; note that A ( i, j, N ) = −∞ if W ( i, j, N ) = ∅ . It is easy to compute A (1 , j, N ) for all j ∈ [0 : k ] and all N ∈ N (1) : we have A (1 , j, N ) = u ( j ) if j = (cid:80) r ∈ [1: m ] n ,r and A (1 , j, N ) = −∞ otherwise. Also, for each i ∈ [1 : n ] we have A ( i, , N ) = 0 if N (cid:96),r = 0 for all (cid:96), r ∈ [1 : m ] and A ( i, , N ) = −∞ otherwise.We will now explain how to compute A ( i, j, N ) given the values of A ( i − , j (cid:48) , N ) for all j (cid:48) ∈ [1 , j ] and all N ∈ N ( i ) . Fix i ∈ [2 : n ] , j ∈ [0 : k ] , N ∈ N ( i ) . Note firstthat for any set W ∈ W ( i, j, N ) we have | v i ∩ W | = (cid:88) r,(cid:96) ∈ [1: m ] n r,(cid:96) ; also, if (cid:80) r,(cid:96) ∈ [1: m ] n r,(cid:96) > j , then W ( i, j, N ) = ∅ .Further, for every set W ∈ W ( i, j, N ) the set W \ { c t | b t = i } belongs to W ( i − , j (cid:48) , N (cid:48) ) for j (cid:48) = j − |{ c t | b t = i }| and for some matrix N (cid:48) ∈ N (cid:48) ( i ) with n (cid:48) (cid:96),r = n (cid:96),r for (cid:96) (cid:54) = i and r (cid:54) = i − . Let j (cid:48) = j − |{ c t | b t = i }| , N (cid:48) = { N (cid:48) ∈N ( i − | n (cid:48) (cid:96),r = n (cid:96),r for (cid:96) (cid:54) = i, r (cid:54) = i − } . Then we have A ( i, j, N ) = max N (cid:48) ∈N (cid:48) ( i,N ) A ( i − , j (cid:48) , N (cid:48) ) + u ( (cid:88) r,(cid:96) ∈ [1: m ] n r,(cid:96) ) if (cid:80) r,(cid:96) ∈ [1: m ] n r,(cid:96) ≤ j and A ( i, j, N ) = −∞ otherwise.To complete the proof, it remains to observe that the PAV -score of an optimal size- k committee is given by max N ∈N ( n ) A ( n, k, N ) . Once this score is computed, therespective committee can be found using standard dynamic programming techniques,and the bound on running time follows immediately.The reader may wonder if constraints on s and d in Theorems 13, 14 and 15 are16ecessary. We conjecture that the answer is yes, i.e., winner determination under PAV remains hard under CI and VI preferences.
Conjecture 17.
PAV is NP-hard even for CI and VI preferences.
However, for “truncated” weight vectors w we can find w - PAV winners in polyno-mial time. As the (1 , , . . . ) - PAV rule is essentially the classic Chamberlin–Courantrule (Chamberlin and Courant, 1983) for dichotomous preferences, our next result canbe seen as an extension of the results of (Betzler et al., 2013) and (Skowron et al.,2015b) for the Chamberlin–Courant rule and single-peaked and single-crossing prefer-ences: while we work on a less expressive domain (dichotomous preferences vs. totalorders), we can handle a larger class of rules (all weight vectors with a constant numberof non-zero entries rather than just (1 , , . . . , ) ). Theorem 18.
Consider a weight vector w where w i = 0 for i > i for some constant i . Then given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if P satisfies VI, we can find a winningcommittee under w - PAV in polynomial time.Proof.
Assume that P satisfies VI with respect to the order of voters v (cid:64) · · · (cid:64) v n .The following algorithm is a refinement of Theorem 13. For each triple ( i, A, (cid:96) ) ,where i ∈ [1 : n ] , A ⊆ v i , and (cid:96) ∈ [0 : k ] , let r ( i, A, (cid:96) ) be the maximum utility that thefirst i voters can obtain from a committee W such that | W | = (cid:96) , and W ⊆ v ∪ . . . ∪ v i and A ⊆ W .We have r (1 , A, (cid:96) ) = u ( (cid:96) ) for every (cid:96) ∈ [0 : | v | ] and A ⊆ v with | A | =min( i , (cid:96) ) . In addition, we have r (1 , A, (cid:96) ) = −∞ for every other A ⊆ v and (cid:96) ∈ [0 : k ] . To compute r ( i + 1 , A, (cid:96) ) for i ∈ [1 : n − , A ⊆ v i +1 with | A | ≤ i and (cid:96) ∈ [ | A | : k ] , we let s = | v i +1 \ ( v i ∪ A ) | , i.e., the maximal number of candidates thatmight have been added in the i + 1 st step to the committee but that do not show up in A , and set r ( i + 1 , A, (cid:96) ) = max r ( i, D ∪ ( A ∩ v i ) , (cid:96) − | A | − r ) + u ( | A | ) , where the maximum is taken over all D ⊆ v i \ v i +1 with | D | ∈ [0 : i − | A ∩ v i | ] andall r ∈ [0 : s ] .This dynamic program has n · m i · ( k + 1) states, and the value of each state iscomputed using O ( m i + 1) arithmetic operations. Assuming that basic calculationstake constant time, we obtain a total runtime of O ( n · m i +1 · k ) , which is polynomialfor constant i . Theorem 19.
Consider a weight vector w where w i = 0 for i > i for some constant i . Then given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if P satisfies CI, we can find a winningcommittee under w - PAV in polynomial time.Proof.
Assume that P satisfies CI with respect to the order of candidates c (cid:67) · · · (cid:67) c m .The following algorithm is a refinement of Theorem 14. For two sets C , C ⊆ C wewrite C (cid:67) C to denote that for all c ∈ C and d ∈ C it holds that c (cid:67) d . For17ach triple ( j, A, (cid:96) ) , where j ∈ [1 : m ] , A ⊆ { c , . . . , c j } , | A | ≤ i and (cid:96) ∈ [0 : k ] ,let r ( j, A, (cid:96) ) be the maximum utility that voters can obtain from a committee W suchthat A ⊆ W , | W | = (cid:96) and W \ A (cid:67) A . Also, for each j ∈ [1 : m − s + 1] andeach A ⊆ { c j , . . . , c j + s − } let t ( A, c ) = (cid:80) v ∈P : c ∈ v u ( | A ∩ v | ) . It is essential that,given a committee W satisfying the conditions above, t ( W, c ) = t ( A, c ) assuming CIpreferences and c (cid:48) (cid:67) c for all c (cid:48) ∈ A \{ c } . Furthermore, note that all the quantities t ( ., . ) can be computed in time O ( n · m i +1 ) since we assume that u ( . ) can be computed inconstant time.We have r (1 , ∅ ,
0) = 0 , r (1 , { c } ,
1) = |{ v i : c ∈ v i }| , and r (1 , A, (cid:96) ) = −∞ if ( A, (cid:96) ) (cid:54) = ( ∅ , , ( { c } , . The quantities r ( j + 1 , A, (cid:96) ) for j ∈ [1 : m − and A ⊆ { c , . . . , c j +1 } with | A | ≤ i can now be computed as follows. If c j +1 (cid:54)∈ A , weset r ( j + 1 , A, (cid:96) ) = r ( j, A, (cid:96) ) . Now, suppose that c j +1 ∈ A . Let A (cid:48) = A \{ c j +1 } . Then r ( j +1 , A, (cid:96) ) = max { r , r }− t ( A (cid:48) , c j +1 ) + t ( A, c j +1 ) where r = max c ∈ C with { c } (cid:67) A r ( j, A (cid:48) ∪ { c } , (cid:96) − ,r = r ( j, A (cid:48) , (cid:96) − . We output max A ⊆ C with | A |≤ i r ( m, A, k ) . Our dynamic program has at most m i +1 · ( k + 1) states, and the utility of each state can be computed in time O ( m ) . Combiningthis with the time used to compute t ( ., . ) , we obtain a total runtime of O ( n · m i +1 · k ) ,which is polynomial for fixed i .Moreover, for the more restricted domains, such as VEI, CEI, WSC and PART wecan design polynomial-time algorithms for both MAV and
PAV , under no additionalconstraints on preferences (again, our results extend to w - PAV ). Theorem 20.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if P satisfies VEI, we can find awinning committee under MAV and
PAV in polynomial time.Proof.
Assume without loss of generality that P satisfies VEI for voter order v (cid:64) · · · (cid:64) v n . Each candidate in C belongs to one of the following four groups: C = v ∩ v n , C = v \ v n , C = v n \ v , and C = v ∩ v n ; candidates in C are approvedby all voters and candidates in C are not approved by any of the voters.Suppose first that | C ∪ C ∪ C | < k . Then there exists an optimal committeefor both PAV and
MAV that contains all candidates in C ∪ C ∪ C and exactly k − | C ∪ C ∪ C | candidates from C . Hence, we can now assume that this is not thecase. Then there exist an optimal committee that contains no candidates from C .Now, if | C | ≥ k , an optimal committee for both PAV and
MAV consists of k candidates from C , and if | C | < k , there exists an optimal committee that containsall candidates in C . It remains to decide how to allocate the remaining places amongcandidates in C and C . To do so, we observe that there is a natural ordering over eachof these sets: given a pair of candidates ( c, c (cid:48) ) in C × C or C × C , we write c ≤ c (cid:48) if { i : c ∈ v i } ⊆ { i : c (cid:48) ∈ v i } . Note that every two candidates in C are comparable18ith respect to ≤ , and so are every two candidates in C . It is now easy to see thatthere exists an optimal committee (for PAV or MAV ) that consists of candidates in C , top p candidates in C with respect to ≤ and top r candidates in C with respect to ≤ for some non-negative values of p, r with p + r + | C | = k . Thus, by considering atmost k possibilities for p and r , we can find an optimal committee.For CEI, we employ a dynamic programing algorithm, somewhat similar to the oneused in Theorem 14. Since we consider a more constrained preferences (CEI insteadof CI), we do not require to maintain an exponential number of states. Theorem 21.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if P satisfies CEI, we can find awinning committee under MAV and
PAV in polynomial time.Proof.
Assume that P satisfies CEI with respect to the order of candidates c (cid:67) · · · (cid:67) c m .Let us consider PAV. For j ∈ [1 : m ] , let V j denote all votes v ∈ P such that c j is therightmost approved candidate of v if c ∈ v and c j is the leftmost approved candidateof v if c / ∈ v . States are identified by a pair ( j, (cid:96) ) , where j ∈ [1 : m ] and (cid:96) ∈ [0 : k ] .Let r ( j, (cid:96) ) be the maximum utility that the voters V ∪ · · · ∪ V j can obtain from acommittee W with | W | = k such that | W ∩ { c , . . . , c j }| = (cid:96) , i.e., W contains (cid:96) candidates to the left of c j (including c j ) and k − (cid:96) candidates to the strictly to the rightof c j .We have r (1 ,
0) = 0 and r (1 ,
1) = | V | , since V contains exactly those votes thatapprove only c . For j ∈ [1 : m ] and (cid:96) ∈ [0 : k ] we have r ( j, (cid:96) ) = −∞ if (cid:96) > j or if k − (cid:96) > m − j . The remaining quantities r ( j +1 , (cid:96) ) for j ∈ [1 : m − can be computedas follows: Let V − j +1 = { v ∈ V j +1 : c j ∈ v } and V + j +1 = { v ∈ V j +1 : c j / ∈ v } . Thequantity r ( j + 1 , (cid:96) ) = max ( r , r ) , where r = r ( j, (cid:96) ) + (cid:88) v ∈ V − j +1 u ( (cid:96) ) + (cid:88) v ∈ V + j +1 u ( k − (cid:96) ) and r = r ( j, (cid:96) −
1) + (cid:88) v ∈ V − j +1 u ( (cid:96) ) + (cid:88) v ∈ V + j +1 u ( k − (cid:96) + 1) . Here, r corresponds to committees that do not contain c j +1 and r to committees thatcontain c j +1 . We output r ( m, k ) . These quantities can be computed in polynomialtime.For MAV we use a similar approach. For j ∈ [1 : m ] and (cid:96) ∈ [0 : k ] , let g ( j, (cid:96) ) bethe minimum MAV-score obtainable by the voters in V ∪ · · · ∪ V j from a committee W with | W | = k such that | W ∩ { c , . . . , c j }| = (cid:96) . Recall that V contains only votes thatapprove c ; hence we have g (1 ,
0) = 0 , g (1 ,
1) = 1 if | V | ≥ and g (1 ,
0) = 1 and g (1 ,
1) = 0 if V = ∅ . For j ∈ [1 : m ] and (cid:96) ∈ [0 : k ] we have g ( j, (cid:96) ) = ∞ if (cid:96) > j or if k − (cid:96) > m − j . The remaining quantities g ( j +1 , (cid:96) ) for j ∈ [1 : m − can be computedas follows: Observe that for v ∈ V − j +1 and a committee W with | W | = k such that | W ∩ { c , . . . , c j }| = (cid:96) , | W \ v | + | v \ W | = ( j + 1 − (cid:96) ) + ( k − (cid:96) ) = j + 1 + k − (cid:96) . For v ∈ V + j +1 , it holds that | W \ v | + | v \ W | = ( m − j − − k + (cid:96) )+( (cid:96) ) = m − ( j +1+ k − (cid:96) ) c j +1 / ∈ v and | W \ v | + | v \ W | = ( m − j − − k + (cid:96) − (cid:96) −
1) = m − ( j +3+ k − (cid:96) ) if c j +1 ∈ v . The quantity g ( j + 1 , (cid:96) ) = min { g , g } , where g = max { r ( j, (cid:96) ) , j + 1 + k − (cid:96), m − ( j + 1 + k − (cid:96) ) } and g = max { r ( j, (cid:96) − , j + 1 + k − (cid:96), m − ( j + 3 + k − (cid:96) ) } . As before, g corresponds to committees that do not contain c j +1 and g to committeesthat contain c j +1 . We output g ( m, k ) . Proposition 22.
Given a dichotomous profile P = ( v , . . . , v n ) over a candidate set C = { c , . . . , c m } and a target committee size k , if P satisfies WSC or PART, we canfind a winning committee under MAV and
PAV in polynomial time.Proof sketch.
For WSC, we can use the characterization in Lemma 1; the problemthen boils down to deciding how many candidates to select from each of the sets u \ w , u ∩ w and w \ u . For PART and PAV , we can show that an optimal committee canbe found by a natural greedy algorithm that at each point selects the candidate withthe largest “marginal contribution” to the total utility. For PART and
MAV , we check,for each t = 0 , . . . , n , whether there exists a committee whose MAV -score is at most t . This is the case if for each voter v ∈ P we can select at least ( | v | + k − t ) / candidates from v . Thus, if v , . . . , v (cid:96) are the distinct votes in P , we need to check that (cid:80) (cid:96)i =1 | v i | ≤ (cid:96)t − ( (cid:96) − k . We have initiated research on analogues of the notions of single-peakedness and single-crossingness for dichotomous preference domains. We have proposed many constraintsthat capture some aspects of what it means for dichotomous preferences to be single-dimensional, explored the relationship among them, and showed that these constraintscan be useful for identifying efficiently solvable special cases of hard voting problemson dichotomous domains. The algorithmic results in Section 4 can be seen as a proofthat our approach has merit; however, there is certainly room for improvement there,both in terms of removing restrictions on the sizes of approval sets and number of votersthat approve each candidate (for
PAV ) and in terms of considering larger domains,such as PSC for
PAV and CI/VI for
MAV .For many of our constraints, we have provided efficient algorithms for check-ing whether a given dichotomous profile satisfies that constraint; only checking ofPSC/SSC remains as an open case. We can also ask if it is possible to detect if agiven dichotomous profile is close to satisfying a structural constraint, and whethersuch “almost-structured” profiles have useful algorithmic properties; similar issues forprofiles of total orders have recently received a lot of attention in the literature (Cornazet al., 2012, 2013; Bredereck et al., 2013; Erd´elyi et al., 2013; Elkind and Lackner,2014; Faliszewski et al., 2014). 20 cknowledgments
This work was supported by the European Research Council (ERC) under grant number639945 (ACCORD). The second author was additionally supported by the AustrianScience Foundation FWF, grant P25518-N23 and Y698.
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