Preference Aggregation on Structured Preference Domains

Edith Elkind University of Oxford

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

ABCD

BCDA

CABD

DABC

BCDA

CDAB

ABCD

BCAD

CABD

Applications

elections

hiringbudget allocation

collaborative filtering

admissions

abcd

abcd

abcd

abcd

abcd

abcd

abcd

mechanism

group decision

paper selection

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

ABCD

BCDA

CABD

DABC

BCDA

CDAB

ABCD

BCAD

CABD

Example: Movie Selection

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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

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 vk+1

Single-Crossing Preferences

Definition: a profile is single-crossing (SC) wrt an ordering of voters (v1, …, vn) if for each pair of candidates A, B there exists an i {0, …, n} such that voters v1, …, vi prefer A to B, and voters vi+1, …, vn prefer B to A

ABCD

BACD

BCAD

CBAD

CBDA

CDBA

DCBA

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 vk+1

– if vk+1 prefers A to B, so do k other voters

ABCD

BACD

BCAD

CBAD

CBDA

CDBA

DCBA

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

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

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• Dswap(x, y): |{(A, B): x prefers A to B, y prefers B to A}|

• Lemma: if u < v < w, then Dswap(u, v) < Dswap(u, w)

…A…B……

……B…A…

…B……A…

u: v: w:

Corollary: SC order is unique (up to reversals and duplicates)

SP SC

Single-Peaked Profile That Is Not Single-Crossing

• v1 and v2 have to be adjacent (because of B, C)

• v3 and v4 have to be adjacent (because of B, C)

• v1 and v3 have to be adjacent (because of A, D)

• v2 and v4 have to be adjacent (because of A, D)

a contradiction

BCAD

BCDA

CBAD

CBDA

DA CB

Single-Crossing Profile That Is Not Single-Peaked

Each candidate is ranked last exactly once

12...……n-2n-1n

nn-1n-2………21

nn-1n-2………12

n12………n-2n-1

nn-112………n-2

…

SP SC

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)

DA CB Ev1 v2v4v3

BACDE

CBDAE

DECBA

EDCBA

1-Euc = SP ∩ SC?

• Proposition [EFS’14, Lackner’14]: There exists a preference profile that is SP and SC, but not 1-Euclidean

v1: 2 3 4 5 1 6v2: 4 5 3 2 1 6 v3: 4 5 6 3 2 1

• SC wrt v1 < v2 < v3, SP wrt 1 < 2 < 3 < 4 < 5 < 6• Not 1-Euclidean: – (x(1)+x(5))/2 < x(v1) < (x(2)+x(3))/2

– (x(3)+x(4))/2 < x(v2) < (x(1)+x(6))/2

– (x(2)+x(6))/2 < x(v3) < (x(4)+x(5))/2

41 32 5 6

SP SC1-Euc

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(c1), …, x(cm), x(v1), …, x(vn)– for each voter v and

each pair of candidates a, b with a < b, if a >v b, add inequality x(v) < (x(a)+x(b))/2, andif b >v a, add inequality x(v) > (x(a)+x(b))/2

Ordering Candidates – 1st 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…

Ordering Candidates – 2nd 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

v1vn

a b

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?

Dichotomous Preferences

• So far, we assumed that votes = orders• What if voters have binary preferences?– voter i approves candidates in Ai,

disapproves candidates in V\Ai

• 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]

Restricted Binary Domains: Examples

• Candidate Interval (CI):candidates can be ordered so that each voter’s approved candidates form an interval

a b d e f gc

u

v

w

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

u v x y z tw

a

b

c

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}

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)

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]

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

Open Problems • Higher dimensions: can we recognize preferences that

are d-Euclidean for d>1 (voters and candidates are points in Rd)?– 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