Computational social choice Lirong Xia Todays schedule - - PowerPoint PPT Presentation

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Computational social choice Lirong Xia Todays schedule - - PowerPoint PPT Presentation

Computational social choice Lirong Xia Todays schedule Computational social choice: the easy-to- compute axiom voting rules that can be computed in P satisfies the axiom Kemeny: a full proof of NP-hardness IP formulation


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Lirong Xia

Computational social choice

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ØComputational social choice: the easy-to- compute axiom

  • voting rules that can be computed in P
  • satisfies the axiom
  • Kemeny: a full proof of NP-hardness
  • IP formulation of Kemeny

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Today’s schedule

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ØKendall tau distance

  • K(R,W)= # {different pairwise comparisons}

ØKemeny(P)=argminW K(P,W) =argminW ΣV∈P K(V,W) ØFor single winner, choose the top-ranked alternative in Kemeny(P)

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Kemeny’s rule

K( b ≻ c ≻ a , a ≻ b ≻ c ) =

2

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Profile P WMG K(P,a ≻ b ≻ c ≻ d)=0+1+2*3+2*5=17

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Example

c d a b 2 2 2 2 2

a ≻ b ≻ c ≻ d b ≻ a ≻ c ≻ d d ≻ a ≻ b ≻ c c ≻ d ≻ b ≻ a

1 1 2 2

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ØFor each linear order W (m! iter)

  • for each vote R in D (n iter)
  • compute K(R,W)

ØFind W* with the smallest total distance

  • W*= argminW K(D,W)=argminW ΣR∈DK(R,W)
  • top-ranked alternative at W* is the winner

ØTakes exponential O(m!n) time!

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Computing the Kemeny winner

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Ø Ranking W → direct acyclic complete graph G(W) Ø Given the WMG G(P) of the input profile P Ø K(P,W) = Σa→b∈G(W)#{V∈P: b ≻ a in V} =Σa→b∈G(W)(n+w(b→a))/2 = nm(m-1)/4 + Σa→b∈G(W) w(b→a)/2 Ø argminW K(P,W)=argminWΣa→b∈G(W) w(b→a) =argminWTotal weight on inconsistent edges in WMG

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Kemeny

a ≻ b ≻ c ≻ d b a c d

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ØTotal weight on inconsistent edges between W and P is: 20

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Example

W= a ≻ b ≻ c ≻ d

b a c d

Profile P:

a b c d

20 16 14 12 8 6

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ØReduction from feedback arc set (FAC)

  • Given a directed graph G and a number k
  • does there exist a way to eliminate no more

than k edges to obtain an acyclic graph?

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Kemeny is NP-hard to compute

d a b c

  • J. Bartholdi III, C. Tovey, M. Trick, Voting schemes for which it can be difficult to tell who

won the election, Social Choice Welfare 6 (1989) 157–165.

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ØThe KendallDistance problem:

  • Given a profile P and a number k,
  • Does there exist a ranking W whose total Kendall

distance is at most k?

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Proof

Instance of FAC Instance of KendallDistance Yes No Yes No P-time d a b c

k

k’=2k+nm(m-1)/4-5

c d a b 2 2 2 2 2 WMG(P): G

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ØFor any edge a→b∈G, define ØPa→b= {a ≻ b ≻ others, Reverse(others) ≻ a ≻ b} ØP = ∪ a→b∈G Pa→b

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Constructing the profile

d a b c WMG(Pa→b)= 2

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ØA voting rule satisfies the easy-to- compute axiom if computing the winner can be done in polynomial time

  • P: easy to compute
  • NP-hard: hard to compute
  • assuming P≠NP

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The easy-to-compute axiom

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ØGiven: a voting rule r ØInput: a preference profile P and an alternative c

  • input size: nmlog m

ØOutput: is c the winner of r under P?

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The winner determination problem

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ØIf following the description of r the winner can be computed in p-time, then r satisfies the easy-to-compute axiom ØPositional scoring rule

  • For each alternative (m iter)
  • for each vote in D (n iter)

– find the position of m, find the score of this position

  • Find the alternative with the largest score (m iter)
  • Total time O(mn+m)=O(mn)

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Computing positional scoring rules

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ØFor each pair of alternatives c,d (m(m-1) iter)

  • let k = 0
  • for each vote V∈P
  • if c>d add 1 to the counter k
  • if d>c subtract 1 from k
  • the weight on the edge c→d is k

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Computing the weighted majority graph

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Satisfiability of easy-to-compute

Rule Complexity

Positional scoring P Plurality w/ runoff STV Copeland Maximin Ranked pairs Kemeny NP-hard Slater Dodgson

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ØFor each pair of alternatives a, b there is a binary variable xab

Øxab = 1 if a>b in W Øxab = 0 if b>a in W

Ømax Σa,bw(a→b)xab

s.t. for all a, b, xab+xba=1 for all a, b, c, xab+xbc+xca≤2

ØDo we need to worry about cycles of >3 vertices? Homework

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Solving Kemeny in practice

No edges in both directions No cycle of 3 vertices

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Manpulation

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Manipulation under plurality rule

(lexicographic tie-breaking)

> > > > > >

> >

Plurality rule Alice Bob Carol

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Strategic behavior (of the agents)

ØManipulation: an agent (manipulator) casts a vote that does not represent her true preferences, to make herself better

  • ff

ØA voting rule is strategy-proof if there is never a (beneficial) manipulation under this rule

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ØVoting time! ØN>M>O à O>M>N

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Using STV?

×4

> > > >

×2

> >

×2

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Any strategy-proof voting rule?

ØNo reasonable voting rule is strategyproof

  • Gibbard-Satterthwaite Theorem [Gibbard Econometrica-73,

Satterthwaite JET-75]: When there are at least three

alternatives, no voting rules except dictatorships satisfy

  • non-imposition: every alternative wins for some profile
  • unrestricted domain: voters can use any linear order

as their votes

  • strategy-proofness

ØRandomized voting rules [Gibbard Econometrica-77] ØMultiple winners [Duggan &Schwartz SCW 00]

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Ø Using Arrow’s impossibility theorem

  • Arrow’s theorem: no ranking rule satisfies non-dictatorship,

universal domain, unanimity, and IIA

ØSuppose G-S does not hold for r (strategy-proof + non-dictatorial), construct a contraduction f to Arrow’s theorem

  • 1. for any profile P and any pair of alternatives (a,b), raise

them to the top-2 to obtain Pab. Let a>b in f(P) iff r(Pab)=a

  • 2. prove that f(P) is a linear order
  • 3. prove that f satisfies non-dictatorship, universal domain,

unanimity, and IIA è contradiction

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Proof idea

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ØRelax non-dictatorship: use a dictatorship ØRestrict the number of alternatives to 2 ØRelax unrestricted domain: mainly pursued by economists

  • Single-peaked preferences

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A few ways out

10% Participation 40% 70%

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ØThere exists a social axis S

  • linear order over the alternatives

ØEach voter’s preferences V are compatible with the social axis S

  • there exists a “peak” a such that
  • [b≺c≺a in S] implies [c≻b in V]
  • [a≻c≻b in S] implies [c≻b in V]
  • alternatives closer to the peak are more preferred
  • different voters may have different peaks, but

must share the same social axis

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Single-peaked preferences

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Examples

rank Axis

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Ø The median rule

  • given a profile of “peaks”
  • choose the median in the social axis

Ø Theorem. The Median rule is strategy-proof. Ø The median rule with phantom voters [Moulin PC 80]

  • parameterized by a fixed set of “peaks” of phantom voters
  • chooses the median of the peaks of the regular voters and

the phantom voters

Ø Theorem. Any strategy-proof rule for single-peaked preferences are median rules with phantom voters

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Strategy-proof rules for single- peaked preferences