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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  1x  s3  1x

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

iB

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



84

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



85

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

86

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









87

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









88

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)



89

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



90

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



91

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)









92


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