Computational social choice Lirong Xia Todays schedule - - PowerPoint PPT Presentation
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
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
2
Today’s schedule
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)
3
Kemeny’s rule
K( b ≻ c ≻ a , a ≻ b ≻ c ) =
2
Profile P WMG K(P,a ≻ b ≻ c ≻ d)=0+1+2*3+2*5=17
4
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
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!
5
Computing the Kemeny winner
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
6
Kemeny
a ≻ b ≻ c ≻ d b a c d
ØTotal weight on inconsistent edges between W and P is: 20
7
Example
W= a ≻ b ≻ c ≻ d
b a c d
Profile P:
a b c d
20 16 14 12 8 6
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?
8
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.
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?
9
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
ØFor any edge a→b∈G, define ØPa→b= {a ≻ b ≻ others, Reverse(others) ≻ a ≻ b} ØP = ∪ a→b∈G Pa→b
10
Constructing the profile
d a b c WMG(Pa→b)= 2
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
11
The easy-to-compute axiom
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?
12
The winner determination problem
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)
13
Computing positional scoring rules
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
14
Computing the weighted majority graph
15
Satisfiability of easy-to-compute
Rule Complexity
Positional scoring P Plurality w/ runoff STV Copeland Maximin Ranked pairs Kemeny NP-hard Slater Dodgson
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
16
Solving Kemeny in practice
No edges in both directions No cycle of 3 vertices
17
Manpulation
Manipulation under plurality rule
(lexicographic tie-breaking)
> > > > > >
> >
Plurality rule Alice Bob Carol
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
ØVoting time! ØN>M>O à O>M>N
20
Using STV?
×4
> > > >
×2
> >
×2
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]
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
22
Proof idea
ØRelax non-dictatorship: use a dictatorship ØRestrict the number of alternatives to 2 ØRelax unrestricted domain: mainly pursued by economists
- Single-peaked preferences
23
A few ways out
10% Participation 40% 70%
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
24
Single-peaked preferences
25
Examples
rank Axis
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
26