1. Preference Aggregation on Structured Preference Domains Edith Elkind University of Oxford

2. Voters and Their Preferences n voters, m candidates Each voter has a complete ranking of the candidates (his preference order) We may want to select: – a single winner – a fixed-size subset of winners (a committee) – a ranking of the candidates ABCDABCD BCDABCDA CABDCABD DABCDABC BCDABCDA CDABCDAB ABCDABCD BCADBCAD CABDCABD

3. Applications elections hiring budget allocation collaborative filtering admissions abcdabcd abcdabcd abcdabcd abcdabcd abcdabcd abcdabcd abcdabcd mechanism group decision paper selection

4. Difficulties Problem: with no assumption on preference structure – counterintuitive behavior may occur majority of voters prefer A to B, B to C, C to A – computational problems are often hard e.g., selecting the most representative committee ABCDABCD BCDABCDA CABDCABD DABCDABC BCDABCDA CDABCDAB ABCDABCD BCADBCAD CABDCABD

5. Example: Movie Selection ?

6. A B C D E F Single-Peaked Preferences Definition: a preference profile is single-peaked (SP) wrt an ordering < of candidates (axis) if for each voter v there exists a candidate C such that: – v ranks C first – if C < D < E, v prefers D to E – if A < B < C, v prefers B to A Example: – voter 1: C > B > D > E > F > A – voter 2: A > B > C > D > E > F – voter 3: E > F > D > C > B > A

7. SP Preferences: Condorcet Winners Claim: in single-peaked elections, the majority relation is transitive Weaker claim: there exists a candidate preferred to every other candidate by a majority of voters (the Condorcet winner) – suppose we have n = 2k+1 voters – order them according to their top choice – pick the top choice of voter v k+1

8. Single-Crossing Preferences Definition: a profile is single-crossing (SC) wrt an ordering of voters (v 1, …, v n ) if for each pair of candidates A, B there exists an i  {0, …, n} such that voters v 1, …, v i prefer A to B, and voters v i+1, …, v n prefer B to A ABCDABCD BACDBACD BCADBCAD CBADCBAD CBDACBDA CDBACDBA DCBADCBA

9. SC Preferences: Majority is Transitive Claim: in single-crossing elections, the majority relation is transitive – suppose we have n=2k+1 voters – consider the ranking of voter v k+1 – if v k+1 prefers A to B, so do k other voters ABCDABCD BACDBACD BCADBCAD CBADCBAD CBDACBDA CDBACDBA DCBADCBA

10. SP and SC Preferences: Algorithms Many NP-hard problems become easy if we assume that preferences are SP or SC Computing Dodgson, Young, and Kemeny winners – coincide with Condorcet winners when they exist Various forms of manipulation [Faliszewski, Hemaspaandra, Hemaspaandra, Rothe’11, …] Computing the most representative committee (Chamberlin-Courant’s rule)[Betzler, Slinko, Uhlmann’13, Skowron, Yu, Faliszewski, E.’13] Computing Plurality election equilibria under random tie-breaking [E., Markakis, Obraztsova, Skowron’?] – dynamic programming algorithms

11. Recognizing SP Preferences It is easy to check whether an election is single-peaked wrt a given axis – but what if the axis is not known? Theorem: SP elections can be recognized in poly-time [Bartholdi, Trick’86, Doignon, Falmagne’94, Escoffier, Lang, Ozturk’08] Observation: if v ranks C last, then C is either the leftmost candidate or the rightmost candidate Corollary: in a SP election at most 2 candidates are ever ranked last

12. Recognizing SC Preferences Theorem: SC elections can be recognized in poly-time [E., Faliszewski, Slinko’12, Bredereck, Chen, Woeginger’12] Theorem’: for each vote u, can decide in poly-time if there is a SC ordering where u appears first D swap (x, y): |{(A, B): x prefers A to B, y prefers B to A}| Lemma: if u < v < w, then D swap (u, v) < D swap (u, w) …A…B………A…B…… ……B…A………B…A… …B……A……B……A… u: v:w: Corollary: SC order is unique (up to reversals and duplicates)

13. SP SC

14. Single-Peaked Profile That Is Not Single-Crossing v 1 and v 2 have to be adjacent (because of B, C) v 3 and v 4 have to be adjacent (because of B, C) v 1 and v 3 have to be adjacent (because of A, D) v 2 and v 4 have to be adjacent (because of A, D) a contradiction BCADBCAD BCDABCDA CBADCBAD CBDACBDA DACB

15. Single-Crossing Profile That Is Not Single-Peaked Each candidate is ranked last exactly once 1 2... … … n-2 n-1 n n n-1 n-2 … … … 2 1 n n-1 n-2 … … … 1 2 n 1 2 … … … n-2 n-1 n n-1 1 2 … … … n-2 …

16. SP SC

17. 1D-Euclidean Preferences Both voters and candidates are points in R v prefers A to B if |v - A| < |v - B| Observation: 1D-Euclidean preferences are – single-peaked (wrt ordering of candidates on the line) – single-crossing (wrt ordering of voters on the line) DACBEv1v1 v2v2 v4v4 v3v3 BACDEBACDE CBDAECBDAE DECBADECBA EDCBAEDCBA

18. 1-Euc = SP ∩ SC? Proposition [EFS’14, Lackner’14]: There exists a preference profile that is SP and SC, but not 1-Euclidean v 1 : 2 3 4 5 1 6 v 2 : 4 5 3 2 1 6 v 3 : 4 5 6 3 2 1 SC wrt v 1 < v 2 < v 3, SP wrt 1 < 2 < 3 < 4 < 5 < 6 Not 1-Euclidean: – (x(1)+x(5))/2 < x(v 1 ) < (x(2)+x(3))/2 – (x(3)+x(4))/2 < x(v 2 ) < (x(1)+x(6))/2 – (x(2)+x(6))/2 < x(v 3 ) < (x(4)+x(5))/2 41325 6

19. SP SC 1-Euc

20. Recognizing 1-Euclidean Preferences Question: can we recognize 1-Euclidean preferences in polynomial time? Observation: if the order of candidates is known, it suffices to solve an LP: – variables x(c 1 ), …, x(c m ), x(v 1 ), …, x(v n ) – for each voter v and each pair of candidates a, b with a v b, add inequality x(v) v a, add inequality x(v) > (x(a)+x(b))/2

21. Ordering Candidates – 1 st Attempt [Knoblauch’10]: there exists a poly-time algorithm for recognizing 1-Euclidean preferences Idea: – check that the input election is SP – if yes, use a SP ordering of the candidates Difficulty: – there can be many SP orderings – some work, others do not An initial SP ordering needs to be tweaked…

22. Ordering Candidates – 2 nd Attempt [EF’14]: there exists a poly-time algorithm for recognizing 1-Euclidean preferences Idea: use the (unique) SC order of voters It works, but… Bad news: this was discovered by Doignon and Falmagne in 1994 v1v1 vnvn ab

23. Eliminating LP and … Observation: when showing that an SPSC profile is not 1-Euclidean, we had a very simple infeasibility certificate Can we identify a simple feasibility criterion that does not involve solving the LP?

24. Dichotomous Preferences So far, we assumed that votes = orders What if voters have binary preferences? – voter i approves candidates in A i, disapproves candidates in V\A i What are the analogues of SP/SC preferences in this setting? Can we recognize the preferences in these restricted domains? Can we exploit then to get efficient algorithms? [E. Lackner, IJCAI’15]

25. Restricted Binary Domains: Examples Candidate Interval (CI): candidates can be ordered so that each voter’s approved candidates form an interval abdefgc u v w

26. Restricted Binary Domains: Examples Voter Interval (VI): voters can be ordered so that for each candidate the set of voters who approve her form an interval uvxyztw a b c

27. Euclidean Preferences Dichotomous Euclidean (DE): voters and candidates can be placed on the line so that for each voter v there is a radius r(v) s.t. v’s approval set is {c: d(c, v) ≤ r(v)} Dichotomous Uniformly Euclidean (DE): voters and candidates can be placed on the line so that there is a radius r s.t. for each voter v his approval set is {c: d(c, v) ≤ r}

28. Refinement-Based Approaches Refinement: a total order a > b > c > d is a refinement of approval vote {a, b} Possibly single-peaked (PSP): there is a single- peaked profile of total orders that is a refinement of the given profile Possibly single-crossing (PSC) Possibly Euclidean (PE)

29. Relationships and Complexity CI = DE = PSP = PE DUE implies CI and VI VI and CI are incomparable VI and CI are easy to detect (consecutive 1s) DUE can be recognized in polynomial time [Nederlof, Woeginger, May’15]

30. Applications PAV: a voting rule to select committees under dichotomous preferences Computing the output of PAV is NP-hard [Aziz et al.’15] – even if each voter approves at most 2 candidates and each candidate is approved by at most 3 voters Our contribution: easiness results for PAV under VI and CI preferences – FPT wrt max size of approval set – XP wrt max number of approvals

31. Open Problems Higher dimensions: can we recognize preferences that are d-Euclidean for d>1 (voters and candidates are points in R d )? – is this problem even in NP? – even for d=2 Trees: can we recognize preferences that are 1-Euclidean on trees (or other median graphs)? Can we decide if a profile can be made 1-Euclidean by deleting k voters or k candidates? – voter deletion: easy for SC, NP-hard for SP – candidate deletion: easy for SP, NP-hard for SC