Computational Aspects of Approval Voting
and Declared-Strategy Voting
A Dissertation Proposal
15 March 2007
Rob LeGrand
Washington University in St. Louis
Computer Science and Engineering
legrand@cse.wustl.edu
Let’s vote!
45 voters 35 voters 20 voters
A B C (1st)
sincere
preferences C C B (2nd)
B A A (3rd)
2
Plurality voting
45 voters 35 voters 20 voters
A B C
sincere
ballots C C B
B A A
A: 45 votes
―zero-information‖
result B: 35 votes
C: 20 votes
3
Plurality voting
45 voters 35 voters 20 voters
A B C
ballots
?
so far C C B
B A A
A: 45 votes
election
state B: 35 votes
C: 0 votes
4
Plurality voting
45 voters 35 voters 20 voters
A B C
strategic
ballots C C B insincerity!
B A A
B: 55 votes
final
[Gibbard ’73]
election A: 45 votes [Satterthwaite ’75]
state
C: 0 votes
5
What is manipulation?
45 voters 35 voters 20 voters
A B C
ballot
sets C C B
BV B A A BU
B: 55 votes
election
state A: 45 votes
C: 0 votes
6
Manipulation decision problem
Existence of Probably Winning Coalition Ballots (EPWCB)
INSTANCE: Set of alternatives A and a distinguished member
a of A; set of weighted cardinal-ratings ballots BV; the
weights of a set of ballots BU which have not been cast;
probability 0 1
QUESTION: Does there exist a way to cast the ballots BU so
that a has at least probability of winning the election with
the ballots BV BU?
• My generalization of problems from the literature:
[Bartholdi, Tovey & Trick ’89] [Conitzer & Sandholm ’02]
[Conitzer & Sandholm ’03]
7
Manipulation decision problem
Existence of Probably Winning Coalition Ballots (EPWCB)
INSTANCE: Set of alternatives A and a distinguished member
a of A; set of weighted cardinal-ratings ballots BV; the
weights of a set of ballots BU which have not been cast;
probability 0 1
QUESTION: Does there exist a way to cast the ballots BU so
that a has at least probability of winning the election with
the ballots BV BU?
• These voters have maximum possible information
– They have all the power (if they have smarts too)
– If this kind of manipulation is hard, any kind is
8
Manipulation decision problem
Existence of Probably Winning Coalition Ballots (EPWCB)
INSTANCE: Set of alternatives A and a distinguished member
a of A; set of weighted cardinal-ratings ballots BV; the
weights of a set of ballots BU which have not been cast;
probability 0 1
QUESTION: Does there exist a way to cast the ballots BU so
that a has at least probability of winning the election with
the ballots BV BU?
• This problem is computationally easy (in P) for:
– plurality voting [Bartholdi, Tovey & Trick ’89]
– approval voting
9
Manipulation decision problem
Existence of Probably Winning Coalition Ballots (EPWCB)
INSTANCE: Set of alternatives A and a distinguished member
a of A; set of weighted cardinal-ratings ballots BV; the
weights of a set of ballots BU which have not been cast;
probability 0 1
QUESTION: Does there exist a way to cast the ballots BU so
that a has at least probability of winning the election with
the ballots BV BU?
• This problem is computationally infeasible (NP-hard) for:
– Hare [Bartholdi & Orlin ’91]
– Borda [Conitzer & Sandholm ’02]
10
What can we do about manipulation?
• One approach: ―tweaks‖ [Conitzer & Sandholm ’03]
– Add an elimination round to an existing protocol
– Drawback: alternative symmetry (―fairness‖) is lost
• What if we deal with manipulation by embracing it?
– Incorporate strategy into the system
– Encourage sincerity as ―advice‖ for the strategy
11
Declared-Strategy Voting
[Cranor & Cytron ’96]
cardinal rational
preferences strategizer
ballot
election
outcome
state
12
Declared-Strategy Voting
[Cranor & Cytron ’96]
sincerity manipulation
cardinal rational
preferences strategizer
ballot
election
outcome
state
• Separates how voters feel from how they vote
• Levels playing field for voters of all sophistications
• Aim: a voter needs only to give honest preferences
13
What is a declared strategy?
A: 0.0
cardinal B: 0.6
preferences
C: 1.0 A: 0
declared voted
strategy
B: 1 ballot
current A: 45 C: 0
election B: 35
state
C: 0
• Captures thinking of a rational voter
14
Can DSV be hard to manipulate?
I propose to show that DSV can be made to be NP-
hard to manipulate (in the EPWCB sense) if a
particular declared strategy is imposed on the
voters.
15
Favorite vs. compromise, revisited
45 voters 35 voters 20 voters
A B C
ballots
?
so far C C B
B A A
A: 45 votes
election
state B: 35 votes
C: 0 votes
16
Approval voting
[Ottewell ’77] [Weber ’77] [Brams & Fishburn ’78]
45 voters 35 voters 20 voters
A B C insincerity
strategic
avoided
ballots C C B
B A A
B: 55 votes
final
election A: 45 votes
state
C: 20 votes
17
Themes of research
• Approval voting systems
• Susceptibility to insincere manipulation
– encouraging sincere ballots
• Effectiveness of various strategies
• Internalizing insincerity
– separating manipulation from the voter
• Complexity issues
– complexity of manipulation
– complexity of calculating the outcome
18
Strands of proposed research
number of outcome Area of research
alternatives
k=1 an approval Voters approve or disapprove a
rating single alternative. What is the
equilibrium approval rating?
k>1 m=1 Voters elect a winner by approval
winner voting. What DSV-style approval
strategies are most effective?
k>1 m≥1 Voters elect a set of alternatives
winners with approval ballots. Which set
most satisfies the least satisfied
voter? [Brams, Kilgour & Sanver ’04]
19
Strands of proposed research
number of outcome Area of research
alternatives
k=1 an approval Voters approve or disapprove a
rating single alternative. What is the
equilibrium approval rating?
k>1 m=1 Voters elect a winner by approval
winner voting. What DSV-style approval
strategies are most effective?
k>1 m≥1 Voters elect a set of alternatives
winners with approval ballots. Which set
most satisfies the least satisfied
voter? [Brams, Kilgour & Sanver ’04]
20
Strands of proposed research
number of outcome Area of research
alternatives
k=1 an approval Voters approve or disapprove a
rating single alternative. What is the
equilibrium approval rating?
k>1 m=1 Voters elect a winner by approval
winner voting. What DSV-style approval
strategies are most effective?
k>1 m≥1 Voters elect a set of alternatives
winners with approval ballots. Which set
most satisfies the least satisfied
voter? [Brams, Kilgour & Sanver ’04]
21
Approval ratings
• Voters are asked about one alternative: Approve or
disapprove?
– like a Presidential approval rating
– typically, average is reported
• Why not allow votes between 0 (full disapproval) and
1 (full approval) and then average them?
– like metacritic.com
• Let’s see what happens when voters are strategic
22
One approach: Average
r 0, .1, .2, .6, .9
v 0, .1, .2, .6, .9
outcome: f avg (v ) .36
.36
0 1
23
One approach: Average
r 0, .1, .2, .6, .9
v 0, .1, .2, 1, .9
outcome: f avg (v ) .44
.44
0 1
24
Another approach: Median
r 0, .1, .2, .6, .9
v 0, .1, .2, .6, .9
outcome: f med (v ) .2
.2
0 1
25
Another approach: Median
r 0, .1, .2, .6, .9
v 0, .1, .2, 1, .9
outcome: f med (v ) .2
.2
0 1
26
Another approach: Median
• Nonmanipulable
– voter i cannot obtain a better result by voting vi ri
– if f med (v ) vi , increasing vi will not change f med (v )
– if f med (v ) vi , decreasing vi will not change f med (v )
• Allows tyranny by a majority
– v 0, 0, 0,1,1,1,1
– f med (v ) 1
– no concession to the 0-voters
27
Average with Declared-Strategy Voting?
• So Median is far from ideal—what now?
– try using Average protocol in DSV context
cardinal rational
preferences strategizer
ballot
election
outcome
state
• But what’s the rational Average strategy?
28
Rational Average strategy
• For 1 i n, voter i should choose vi to move
outcome as close to ri as possible
• Choosing vi ri n j i v j would give f avg (v ) ri
• Optimal vote is vi min(max(ri n j i v j , 0),1)
• After voter i uses this strategy, one of these is true:
– f avg (v ) ri and vi 1
– f avg (v ) ri
– f avg (v ) ri and vi 0
29
Multiple equilibria are possible
r .2, .3, .5, .5, .8
v 0, 0, .5, 1, 1
v 0, 0, .6, .9, 1
v 0, 0, .75, .75, 1
outcome in each case:
f avg (v ) .5
Multiple equilibria always have same average
(proof in written proposal)
30
An equilibrium always exists?
• At equilibrium, v must satisfy
(i) vi min(max(ri n j i v j , 0),1)
I propose to prove that, given a vector r , at least
one equilibrium exists.
• If an equilibrium always exists, then average at
equilibrium can be defined as a function, f aveq (r ) .
• Applying f aveq to v instead of r gives a new
system, Average-approval-rating DSV.
31
Average-approval-rating DSV
r 0, .1, .2, .6, .9
v 0, .1, .2, .6, .9
outcome: f aveq (v ) .4
.4
0 1
32
Average-approval-rating DSV
r 0, .1, .2, .6, .9
v 0, .1, .2, 1, .9
outcome: f aveq (v ) .4
.4
0 1
33
Average-approval-rating DSV
• AAR DSV could be manipulated if some voter i
could gain an outcome closer to ideal by voting
insincerely ( vi ri ).
I propose to show that Average-approval-rating
DSV cannot be manipulated by insincere voters.
34
Average-approval-rating DSV
• AAR DSV could be manipulated if some voter i
could gain an outcome closer to ideal by voting
insincerely ( vi ri ).
I propose to show that Average-approval-rating
DSV cannot be manipulated by insincere voters.
• Intuitively, if f aveq (v ) vi , increasing vi will not
change f aveq (v ).
35
Higher-dimensional outcome space
• What if votes and outcomes exist in d 1
dimensions?
• Example: x, y 2 : 0 x 1 0 y 1
• If dimensions are independent, Average, Median
and Average-approval-rating DSV can operate
independently on each dimension
– Results from one dimension transfer
36
Higher-dimensional outcome space
• But what if the dimensions are not independent?
– say, outcome space is a disk in the plane:
x, y : x2 y2 1
2
• A generalization of Median: the Fermat-Weber point
[Weber ’29]
– minimizes sum of Euclidean distances between outcome
point and voted points
– F-W point is computationally infeasible to calculate
exactly [Bajaj ’88] (but approximation is easy [Vardi ’01])
– cannot be manipulated by moving a voted point directly
away from the F-W point [Small ’90]
37
Higher-dimensional outcome space
• Average-approval-rating DSV can be generalized
– optimal strategy moves the result as close to sincere
ideal as possible (by Euclidean distance)
I propose to find the optimal strategy for Average in
the x, y 2 : x 2 y 2 1 case and determine
whether the resulting DSV system is rotationally
invariant and/or nonmanipulable by insincere
voters.
38
Strands of proposed research
number of outcome Area of research
alternatives
k=1 an approval Voters approve or disapprove a
rating single alternative. What is the
equilibrium approval rating?
k>1 m=1 Voters elect a winner by approval
winner voting. What DSV-style approval
strategies are most effective?
k>1 m≥1 Voters elect a set of alternatives
winners with approval ballots. Which set
most satisfies the least satisfied
voter? [Brams, Kilgour & Sanver ’04]
39
Approval strategies for DSV
• Rational plurality strategy has been well explored
[Cranor & Cytron, ’96]
• But what about approval strategy?
• If each alternative’s probability of winning is known,
optimal strategy can be computed [Merrill ’88]
• But what about in a DSV context?
– have only a vote total for each alternative
• Let’s look at several approval strategies and
approaches to evaluating their effectiveness
40
DSV-style approval strategies
• Strategy Z [Merrill ’88]:
– Approve alternatives with higher-than-average cardinal
preference (zero-information strategy)
s [30, 25,15,10]
p [0, 1, .8, .3]
Z recommends:
b [0, 1, 1, 0]
41
DSV-style approval strategies
• Strategy T [Ossipoff ’02]:
– Approve favorite of top two vote-getters, plus all liked
more
s [30, 25,15,10]
p [0, 1, .8, .3]
T recommends:
b [0, 1, 0, 0]
42
DSV-style approval strategies
• Strategy J [Brams & Fishburn ’83]:
– Use strategy Z if it distinguishes between top two vote-
getters; otherwise use strategy T
s [30, 25,15,10]
p [0, 1, .8, .3]
J recommends:
b [0, 1, 1, 0]
43
DSV-style approval strategies
• Strategy A:
– Approve all preferred to top vote-getter, plus top vote-
getter if preferred to second-highest vote-getter
s [30, 25,15,10]
p [0, 1, .8, .3]
A recommends:
b [0, 1, 1, 1]
But how to evaluate these strategies?
44
Election-state-evaluation approaches
• Evaluate a declared strategy by evaluating the
election states that are immediately obtained
• Calculate expected value of an election state by
estimating each alternative’s probability of
eventually winning
• How to calculate those probabilities?
45
Election-state-evaluation:
Merrill metric
• Estimate an alternative’s probability of winning to
be proportional to its current vote total raised to
some power x [Merrill ’88]
x
s
w i
i k
sj
j 1
46
Strategy comparison using the Merrill metric
Current election state s [ s1 , s2 , s3 ] s1 s2 s3
Focal voter’s preferences p [ p1 , p2 , p3 ]
p1 p2 p3 [1, 0, 0] (strategies A & T)
p1 p3 p2 [1, 0, 0] (A & T)
p2 p1 p3 [0, 1, 0] (A & T)
p2 p3 p1 [0, 1, 1] (A); [0, 1, 0] (T)
p3 p1 p2 [1, 0, 1] (A & T)
p3 p2 p1 [0, 1, 1] (A & T)
47
Strategy comparison using the Merrill metric
Current election state s [ s1 , s2 , s3 ] s1 s2 s3
Focal voter’s preferences p [ p1 , p2 , p3 ] p2 p3 p1
expected values of possible next election states:
p1s1x p2 s2 1 p3 s3 1
x x
V[ 0,1,1] [0, 1, 1] (A)
s1
x
s2 1x s3 1x
p1s1x p2 s2 1 p3 s3x
x
V[ 0,1,0] [0, 1, 0] (T)
s1x s2 1 s3x
x
48
Strategy comparison using the Merrill metric
Current election state s [ s1 , s2 , s3 ] s1 s2 s3
Focal voter’s preferences p [ p1 , p2 , p3 ] p2 p3 p1
so T is better than A only when:
p1s1x p2 s2 1 p3 s3 1 p1s1x p2 s2 1 p3 s3
x x x x
s1 s2 1 s3 1 s1x s2 1 s3
x x x x x
or, equivalently:
x
p2 p3 s1
s 1
p3 p1 2
49
Strategy comparison using the Merrill metric
Current election state s [ s1 , s2 , s3 ] s1 s2 s3
Focal voter’s preferences p [ p1 , p2 , p3 ] p2 p3 p1
so T is better than A only when:
p1s1x p2 s2 1 p3 s3 1 p1s1x p2 s2 1 p3 s3
x x x
x
s1 s2 1 s3 1 s1x s2 1 s3
x x x x x
or, equivalently:
x
p2 p3 s1 Intuitively, T does better than A only when:
s 1
• s1 and s2 are relatively close
p3 p1 2 • x is relatively small
• p3 is relatively close to p1 compared to p2
50
Strategy comparison using the Merrill metric
Current election state s [ s1 , s2 , s3 ] s1 s2 s3
Focal voter’s preferences p [ p1 , p2 , p3 ] p2 p3 p1
x
T is better than A only when:
p2 p3 s1
s 1
p3 p1 2
Corollaries:
– When x is taken to infinity and s1 s2 1, strategy A
dominates strategy T
– When p1 p2
p3 , strategy A dominates strategy T
2
51
Approval strategy evaluation
I propose to extend this 3-alternative result to
strategy pairs A vs. J, T vs. J and A vs. Z.
I propose to extend this result to strategy pairs A
vs. T and A vs. J in the 4-alternative case.
52
Further result for strategy A
More generally, it is true that if
– the election state is free of ties and near-ties:
s1 s2 1 s3 2 sk k 1
– and the focal voter’s cardinal preferences are tie-free:
pi p j when i j
– and the Merrill-metric exponent x is taken to infinity
then strategy A dominates all other strategies
according to the Merrill metric
• (proof in written proposal)
53
Election-state-evaluation:
Branching-probabilities metric
• Estimate an alternative’s probability of winning by looking
ahead
• Assume that the probability that alternative a is approved on
each future ballot is equal to the proportion of already-voted
ballots that approve a
p1
p2 k p
iB
i 1
p2
54
Approval strategy evaluation
I propose to extend the Merrill-metric results to
strategy pairs A vs. T, A vs. J, T vs. J and A vs. Z in
the 3-alternative case using the branching-
probabilities metric.
I propose to determine whether strategy A
dominates all others in the near-tie-free case using
the branching-probabilities metric as the number of
future ballots goes to infinity.
55
Strands of proposed research
number of outcome Area of research
alternatives
k=1 an approval Voters approve or disapprove a
rating single alternative. What is the
equilibrium approval rating?
k>1 m=1 Voters elect a winner by approval
winner voting. What DSV-style approval
strategies are most effective?
k>1 m≥1 Voters elect a set of alternatives
winners with approval ballots. Which set
most satisfies the least satisfied
voter? [Brams, Kilgour & Sanver ’04]
56
Electing a committee from approval ballots
approves of
k = 5 candidates 11110 00011 candidates
4 and 5
n = 6 ballots
01111 00111
10111 00001
•What’s the best committee of size m = 2?
57
Sum of Hamming distances
m = 2 winners 11110 00011
2 4
4 5
01111 11000 00111
4 3 sum = 22
10111 00001
58
Fixed-size minisum
m = 2 winners 11110 00011
4 0
2 1
01111 00011 00111
2 1 sum = 10
10111 00001
•Minisum elects winner set with smallest sumscore
•Easy to compute (pick candidates with most approvals)
59
Maximum Hamming distance
m = 2 winners 11110 00011
4 0
2 1
01111 00011 00111
2 1 sum = 10
max = 4
10111 00001
60
Fixed-size minimax
[Brams, Kilgour & Sanver ’04]
m = 2 winners 11110 00011
2 2
2 1
01111 00110 00111
2 3 sum = 12
max = 3
10111 00001
•Minimax elects winner set with smallest maxscore
•Harder to compute?
61
Complexity
Endogenous minimax Bounded-size minimax Fixed-size minimax
= EM = BSM(0, k) = BSM(m1, m2) = FSM(m) = BSM(m, m)
NP-hard NP-hard
?
[Frances & Litman ’97] (generalization of EM)
62
Complexity
Endogenous minimax Bounded-size minimax Fixed-size minimax
= EM = BSM(0, k) = BSM(m1, m2) = FSM(m) = BSM(m, m)
NP-hard NP-hard NP-hard
[Frances & Litman ’97] (generalization of EM) (proof in written
proposal)
63
Approximability
Endogenous minimax Bounded-size minimax Fixed-size minimax
= EM = BSM(0, k) = BSM(m1, m2) = FSM(m) = BSM(m, m)
has a PTAS* no known PTAS no known PTAS
[Li, Ma & Wang ’99]
* Polynomial-Time Approximation Scheme: algorithm
with approx. ratio 1 + ε that runs in time polynomial in
the input and exponential in 1/ε
64
Approximability
Endogenous minimax Bounded-size minimax Fixed-size minimax
= EM = BSM(0, k) = BSM(m1, m2) = FSM(m) = BSM(m, m)
has a PTAS* no known PTAS; no known PTAS;
has a 3-approx. has a 3-approx.
[Li, Ma & Wang ’99]
(proof in written (proof in written
proposal) proposal)
* Polynomial-Time Approximation Scheme: algorithm
with approx. ratio 1 + ε that runs in time polynomial in
the input and exponential in 1/ε
65
Approximating FSM
11110 m = 2 winners
00011
00111
00111
00001 choose
a ballot
10111
arbitrarily
01111
66
Approximating FSM
11110 m = 2 winners
00011
00111
coerce to
00111 00101
00001 size m
choose
a ballot
10111
arbitrarily
01111
outcome =
m-completed ballot
67
Approximation ratio ≤ 3
optimal
11110
2 FSM set
00011 2
00111 1
00110
3
00001
2
10111
2
01111
≤ OPT
OPT = optimal maxscore
68
Approximation ratio ≤ 3
optimal chosen
11110
2 FSM set ballot
00011 2
00111 1
1
00110 00111
3
00001
2
10111
2
01111
≤ OPT ≤ OPT
OPT = optimal maxscore
69
Approximation ratio ≤ 3
optimal chosen m-completed
11110
2 FSM set ballot ballot
00011 2
00111 1
1 1
00110 00111 00011
3
00001
2
10111
2
01111
≤ OPT ≤ OPT ≤ OPT
(by triangle inequality)
OPT = optimal maxscore ≤ 3·OPT
70
Better in practice?
• So far, we can guarantee a winner set no more than 3 times
as bad as the optimal.
– Nice in theory . . .
• How can we do better in practice?
– Try local search
71
Local search approach for FSM
1. Start with some c {0,1}k
of weight m
01001
4
72
Local search approach for FSM
1. Start with some c {0,1}k
of weight m
11000 10001
2. In c, swap up to r 0-bits 5 4
with 1-bits in such a way
01100 01001 00101
that minimizes the 4 4 4
maxscore of the result
01010 00011
4 4
73
Local search approach for FSM
1. Start with some c {0,1}k
of weight m
2. In c, swap up to r 0-bits
with 1-bits in such a way
that minimizes the
maxscore of the result
01010
4
74
Local search approach for FSM
1. Start with some c {0,1}k
of weight m
2. In c, swap up to r 0-bits
with 1-bits in such a way
01010
that minimizes the 4
maxscore of the result
75
Local search approach for FSM
1. Start with some c {0,1}k
of weight m
11000 10010
2. In c, swap up to r 0-bits 5 4
with 1-bits in such a way
01100 01010 00110
that minimizes the 4 4 3
maxscore of the result
01001 00011
3. Repeat step 2 until 4 4
maxscore(c) is
unchanged k times
4. Take c as the solution
76
Local search approach for FSM
1. Start with some c {0,1}k
of weight m
2. In c, swap up to r 0-bits
with 1-bits in such a way
00110
that minimizes the 3
maxscore of the result
3. Repeat step 2 until
maxscore(c) is
unchanged k times
4. Take c as the solution
77
Heuristic evaluation
• Parameters:
– starting point of search
– radius of neighborhood
• Ran heuristics on generated and real-world data
• All heuristics perform near-optimally
– highest approx. ratio found: 1.2 (maxscore of solution found)
– highest average ratio < 1.04 (maxscore of exact solution)
• The fixed-size-minisum starting point performs best overall
(with our 3-approx. a close second)
• When neighborhood radius is larger, performance improves
and running time increases
78
Manipulating FSM
00110 00011 m = 2 winners
2 0
2 1
01111 00011 00111
2 1
max = 2
10111 00001
•Voters are sincere
•Another optimal solution: 00101 79
Manipulating FSM
00110
11110 00011 m = 2 winners
0 2 2
2 1
01111 00110 00111
2 3
max = 3
10111 00001
•A voter manipulates and realizes ideal outcome
•But our 3-approximation for FSM is nonmanipulable
80
Fixed-size Minimax contributions
• BSM and FSM are NP-hard
• Both can be approximated with ratio 3
• Polynomial-time local search heuristics perform
well in practice
– some retain ratio-3 guarantee
• Exact FSM can be manipulated
• Our 3-approximation for FSM is nonmanipulable
81
Progress so far
Area of research State of progress
Approval rating Completed: rational Average strategy, equality of
average at equilibria
To do: equilibrium always exists, nonmanipulability of
AAR DSV, analysis of Average in planar disk
DSV-style Completed: comparison of A and T in 3-alt. case,
approval domination of A as x
strategies To do: comparisons of other pairs, analysis using
branching-probabilities metric
Fixed-size Completed: NP-hardness proof, 3-approximation,
minimax heuristic evaluation, manipulability analysis
82
Fin
Thanks to
– my adviser, Ron Cytron
– Morgan Deters and the rest of the DOC Group
– co-authors Vangelis Markakis and Aranyak Mehta
– my committee
Questions?
83
What happens at equilibrium?
• The optimal strategy recommends that no voter
change
• So (i ) v ri vi 1
• And (i ) v ri vi 0
– equivalently, (i ) vi 0 v ri
• Therefore any average at equilibrium must satisfy
two equations:
– (A) v n i : v ri
– (B) i : v ri vn
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Proof: Only one equilibrium average
A( ) n i : ri
B( ) i : ri n
• Theorem:
A(1 ) B(1 ) A(2 ) B(2 ) 1 2
• Proof considers two symmetric cases:
– assume 1 2
– assume 2 1
• Each leads to a contradiction
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Proof: Only one equilibrium average
case 1: 1 2
(i ) 2 ri 1 ri
i : 2 ri i : 1 ri
i : 2 ri i : 1 ri
2n i : 2 ri A(2 )
i : 1 ri 1n B (1 )
2n i : 2 ri i : 1 ri 1n
2 n 1n
2 1 , contradicting 1 2
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Proof: Only one equilibrium average
Case 1 shows that 1 2
Case 2 is symmetrical and shows that 2 1
Therefore 1 2
Therefore, given r , the average at equilibrium is unique
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Specific FSM heuristics
• Two parameters:
– where to start vector c:
1. a fixed-size-minisum solution
2. a m-completion of a ballot (3-approx.)
3. a random set of m candidates
4. a m-completion of a ballot with highest maxscore
– radius of neighborhood r: 1 and 2
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Heuristic evaluation
• Real-world ballots from GTS 2003 council election
• Found exact minimax solution
• Ran each heuristic 5000 times
• Compared exact minimax solution with heuristics to find
realized approximation ratios
– example: 15/14 = 1.0714
• maxscore of solution found = 15
• maxscore of exact solution = 14
• We also performed experiments using ballots generated
according to random distributions (see paper)
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Average approx. ratios found
radius = 1 radius = 2
fixed-size 1.0012 1.0000
minimax
3-approx. 1.0017 1.0000
random 1.0057 1.0000
set
highest- 1.0059 1.0000
maxscore
performance on GTS ’03 election data
k = 24 candidates, m = 12 winners, n = 161 ballots
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Largest approx. ratios found
radius = 1 radius = 2
fixed-size 1.0714 1.0000
minimax
3-approx. 1.0714 1.0000
random 1.0714 1.0000
set
highest- 1.0714 1.0000
maxscore
performance on GTS ’03 election data
k = 24 candidates, m = 12 winners, n = 161 ballots
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Conclusions from all experiments
• All heuristics perform near-optimally
– highest ratio found: 1.2
– highest average ratio < 1.04
• When radius is larger, performance improves and running
time increases
• The fixed-size-minisum starting point performs best overall
(with our 3-approx. a close second)
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