Structured P reference 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 ID: 362200
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Slide1
Preference Aggregation on
Structured
P
reference Domains
Edith Elkind
University of
OxfordSlide2
Voters and Their Preferencesn voters,
m
candidates
Each voter has a complete ranking of the candidates (his preference order)We may want to select:a single winnera fixed-size subset of winners (a committee)a ranking of the candidates
ABCD
BCDA
CABD
DABC
BCDA
CDAB
ABCD
BCAD
CABDSlide3
Applications
elections
hiring
budget
allocation
collaborative filtering
admissions
abcd
abcd
abcd
abcd
abcd
abcd
abcd
mechanism
group decision
paper selectionSlide4
DifficultiesProblem:
with no assumption on preference structure
counterintuitive behavior
may occurmajority of voters prefer A to B, B to C, C to Acomputational problems are often hard e.g., selecting the most representative committee
ABCD
BCDA
CABD
DABC
BCDA
CDAB
ABCD
BCAD
CABDSlide5
Example: Movie Selection
?Slide6
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 Eif A < B < C, v prefers B to AExample: voter 1: C > B > D > E > F > A voter 2: A > B > C > D > E > Fvoter 3: E > F > D > C > B > A Slide7
SP Preferences: Condorcet Winners Claim: in
single-peaked
elections,
the majority relation is transitiveWeaker claim: there exists a candidate preferred to every other candidate by a majority of voters (the Condorcet winner)suppose we have n = 2k+1 votersorder them according to their top choicepick the top choice of voter vk+1Slide8
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, …, v
i prefer A to B
, and voters vi+1
, …, vn prefer
B to A
AB
CD
BACD
B
C
A
DCBADCBDACDBADCBASlide9
SC Preferences: Majority is TransitiveClaim: in
single-crossing
elections,
the majority relation is transitivesuppose we have n=2k+1 votersconsider the ranking of voter vk+1if vk+1 prefers A to B, so do k
other voters
AB
CD
BACD
B
CAD
CB
AD
CBDA
C
D
B
ADCBASlide10
SP and SC Preferences: AlgorithmsMany
NP-hard
problems become easy if we assume that preferences are
SP or SCComputing Dodgson, Young, and Kemeny winnerscoincide with Condorcet winners when they existVarious 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 Slide11
Recognizing SP Preferences It is
easy
to check whether an election is
single-peaked wrt a given axisbut what if the axis is not known? Theorem: SP elections can be recognized in poly-time [Bartholdi, Trick’86, Doignon, Falmagne’94, Escoffier
, L
ang, Ozturk’08]Observation: if v ranks
C last, then C is either the leftmost candidate or the
rightmost candidateCorollary: in a SP
election at most 2 candidates are ever ranked last Slide12
Recognizing SC PreferencesTheorem:
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 firstD
swap(
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)Slide13
SP
SCSlide14
Single-Peaked Profile That Is Not Single-Crossingv
1
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)v
2 and v4
have to be adjacent (because of A, D) a contradiction
BC
AD
BCDA
CBAD
CB
DA
D
ACBSlide15
Single-Crossing Profile That Is Not Single-PeakedEach candidate is ranked last
exactly once
1
2...……n-2
n-1
n
nn-1n-2
……
…2
1
nn-1
n-2…
……
1
2
n
12………n-2n-1nn-112………n-2…Slide16
SP
SCSlide17
1D-Euclidean PreferencesBoth
voters a
nd
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)
D
A
C
B
E
v
1
v
2v4v3BACDECBDAEDECBAEDCBASlide18
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 6
v2:
4 5 3 2 1 6 v3:
4 5 6 3 2 1
SC wrt v
1 < v2 < v
3, SP wrt
1 < 2 < 3 < 4 < 5 < 6Not 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 413256Slide19
SP
SC
1-EucSlide20
Recognizing 1-Euclidean PreferencesQuestion: 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))/2Slide21
Ordering Candidates – 1st Attempt
[Knoblauch’10]
: there exists a poly-time algorithm for recognizing
1-Euclidean preferencesIdea: check that the input election is SPif yes, use a SP ordering of the candidatesDifficulty:there can be many SP orderingssome work, others do notAn initial SP ordering needs to be
tweaked…Slide22
Ordering Candidates – 2nd Attempt
[EF’14]
: there exists a poly-time algorithm for recognizing
1-Euclidean preferencesIdea: use the (unique) SC order of voters It works, but…Bad news: this was discovered by Doignon and Falmagne in 1994
v
1
v
n
a
bSlide23
Eliminating LP and … Observation
: when showing that an
SP
SC profile is not 1-Euclidean, we had a very simple infeasibility certificateCan we identify a simple feasibility criterion that does not involve solving the LP?Slide24
Dichotomous PreferencesSo far, we assumed that votes = ordersWhat if voters have binary preferences?
voter
i
approves candidates in Ai, disapproves candidates in V\AiWhat 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]Slide25
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
g
c
u
v
wSlide26
Restricted Binary Domains: ExamplesVoter 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
t
w
a
b
cSlide27
Euclidean PreferencesDichotomous 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}Slide28
Refinement-Based ApproachesRefinement: 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)Slide29
Relationships and ComplexityCI = DE = PSP = PEDUE implies CI and VIVI and CI are incomparable
VI and CI are easy to detect (consecutive 1s)
DUE can be recognized in polynomial time
[Nederlof, Woeginger, May’15]Slide30
ApplicationsPAV: a voting rule to select committees under dichotomous preferencesComputing 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 setXP wrt max number of approvalsSlide31
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=2Trees: 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
SPcandidate deletion: easy for SP, NP-hard for
SC