computational aspects of approval voting and declared-strategy voting rob legrand washington...

Computational Aspects of Approval Voting

and Declared-Strategy Voting

Rob LeGrandWashington University in St. Louis

Computer Science and Engineeringlegrand@cse.wustl.edu

A Dissertation Proposal15 March 2007

Let’s vote!

45 voters

sincere

preferences

35 voters

20 voters

Plurality voting

A: 45 votes

B: 35 votes

C: 20 votes

sincere

ballots

45 voters

35 voters

20 voters

“zero-information”

result

Plurality voting

ballots

so far

election

45 voters

35 voters

A: 45 votes

B: 35 votes

C: 0 votes

20 voters

Plurality voting

B: 55 votes

A: 45 votes

C: 0 votes

strategic

ballots

election

45 voters

35 voters

20 voters

[Gibbard ’73] [Satterthwaite ’75]

insincerity!

What is manipulation?

B: 55 votes

A: 45 votes

C: 0 votes

45 voters

35 voters

20 voters

A BUBV

ballot

election

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

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 ?

• My generalization of problems from the literature: [Bartholdi, Tovey & Trick ’89] [Conitzer & Sandholm ’02]

[Conitzer & Sandholm ’03]

• 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

• This problem is computationally easy (in P) for:– plurality voting [Bartholdi, Tovey & Trick ’89]

– approval voting

• This problem is computationally infeasible (NP-hard) for:– Hare [Bartholdi & Orlin ’91]

– Borda [Conitzer & Sandholm ’02]

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

Declared-Strategy Voting[Cranor & Cytron ’96]

electionstate

cardinal

preferences

rational

strategizer

ballot

outcome

Declared-Strategy Voting[Cranor & Cytron ’96]

electionstate

cardinal

preferences

rational

strategizer

ballot

outcome

• 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

sincerity manipulation

What is a declared strategy?

A: 0.0

B: 0.6

C: 1.0

cardinal

preferences

current

election

declared

strategy

ballot

• Captures thinking of a rational voter

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.

Favorite vs. compromise, revisited

ballots

so far

election

45 voters

35 voters

A: 45 votes

B: 35 votes

C: 0 votes

20 voters

Approval voting[Ottewell ’77] [Weber ’77] [Brams & Fishburn ’78]

strategic

ballots

45 voters

35 voters

20 voters

B: 55 votes

A: 45 votes

C: 20 votes

election

insincerityavoided

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

Strands of proposed research

number of alternatives

outcome Area of research

k = 1 an approval rating

Voters approve or disapprove a single alternative. What is the equilibrium approval rating?

k > 1 m = 1 winner

Voters elect a winner by approval voting. What DSV-style approval strategies are most effective?

k > 1 m ≥ 1 winners

Voters elect a set of alternatives with approval ballots. Which set most satisfies the least satisfied voter? [Brams, Kilgour & Sanver ’04]

k > 1 m = 1 winner

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

One approach: Average

9.,6.,2.,1.,0v

9.,6.,2.,1.,0r

0 136.

outcome: 36.)( vfavg

One approach: Average

0 144.

9.,1,2.,1.,0v

9.,6.,2.,1.,0r

outcome: 44.)( vfavg

Another approach: Median

9.,6.,2.,1.,0v

9.,6.,2.,1.,0r

outcome: 2.)( vfmed

9.,1,2.,1.,0v

9.,6.,2.,1.,0r

outcome: 2.)( vfmed

• Nonmanipulable– voter i cannot obtain a better result by voting– if , increasing will not change– if , decreasing will not change

• Allows tyranny by a majority– – – no concession to the 0-voters

ii rv imed vvf )(

imed vvf )( iv

1,1,1,1,0,0,0v

1)( vfmed

)(vfmed

Average with Declared-Strategy Voting?

electionstate

cardinal

preferences

rational

strategizer

ballot

outcome

• So Median is far from ideal—what now?– try using Average protocol in DSV context

• But what’s the rational Average strategy?

Rational Average strategy

• For , voter i should choose to move outcome as close to as possible

• Choosing would give• Optimal vote is

• After voter i uses this strategy, one of these is true:– and– – and

ni 1iv

)1),0,min(max(

ij jii vnrviavg rvf )(

ij jii vnrv

iavg rvf )(

0iviavg rvf )(

iavg rvf )(

Multiple equilibria are possible

outcome in each case:

5.)( vfavg

1,1,5.,0,0v 8.,5.,5.,3.,2.r

1,9.,6.,0,0v

1,75.,75.,0,0v

Multiple equilibria always have same average(proof in written proposal)

An equilibrium always exists?

• At equilibrium, must satisfy

I propose to prove that, given a vector , at least one equilibrium exists.

• If an equilibrium always exists, then average at equilibrium can be defined as a function, .

• Applying to instead of gives a new system, Average-approval-rating DSV.

)1),0,min(max()(

ij jii vnrviv

)(rfaveq

Average-approval-rating DSV

9.,6.,2.,1.,0v

9.,6.,2.,1.,0r

outcome: 4.)( vfaveq

9.,1,2.,1.,0v

9.,6.,2.,1.,0r

outcome: 4.)( vfaveq

• AAR DSV could be manipulated if some voter i could gain an outcome closer to ideal by voting insincerely ( ).

I propose to show that Average-approval-rating DSV cannot be manipulated by insincere voters.

• AAR DSV could be manipulated if some voter i could gain an outcome closer to ideal by voting insincerely ( ).

I propose to show that Average-approval-rating DSV cannot be manipulated by insincere voters.

• Intuitively, if , increasing will not change .

iaveq vvf )(

iv)(vfaveq

Higher-dimensional outcome space

• What if votes and outcomes exist in dimensions?

• Example:• If dimensions are independent, Average, Median

and Average-approval-rating DSV can operate independently on each dimension– Results from one dimension transfer

1010:, 2 yxyx

• But what if the dimensions are not independent?– say, outcome space is a disk in the plane:

• 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]

1:, 222 yxyx

• 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 case and determine whether the resulting DSV system is rotationally invariant and/or nonmanipulable by insincere voters.

1:, 222 yxyx

k > 1 m = 1 winner

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

DSV-style approval strategies

• Strategy Z [Merrill ’88]:– Approve alternatives with higher-than-average cardinal

preference (zero-information strategy)

]10,15,25,30[s

]3.,8.,1,0[p

]0,1,1,0[b Z recommends:

• Strategy T [Ossipoff ’02]:– Approve favorite of top two vote-getters, plus all liked

]10,15,25,30[s

]3.,8.,1,0[p

]0,0,1,0[b T recommends:

• Strategy J [Brams & Fishburn ’83]:– Use strategy Z if it distinguishes between top two vote-

getters; otherwise use strategy T

]10,15,25,30[s

]3.,8.,1,0[p

]0,1,1,0[b J recommends:

• Strategy A:– Approve all preferred to top vote-getter, plus top vote-

getter if preferred to second-highest vote-getter

But how to evaluate these strategies?

]10,15,25,30[s

]3.,8.,1,0[p

]1,1,1,0[bA recommends:

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?

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]

Strategy comparison using the Merrill metric

],,[ 321 ssss

321 sss ],,[ 321 pppp

Current election state

Focal voter’s preferences

321 ppp 231 ppp

312 ppp

132 ppp

213 ppp

123 ppp

[1, 0, 0] (strategies A & T)

[1, 0, 0] (A & T)

[0, 1, 0] (A & T)

[0, 1, 1] (A); [0, 1, 0] (T)

[1, 0, 1] (A & T)

[0, 1, 1] (A & T)

],,[ 321 ssss

321 sss ],,[ 321 pppp

Focal voter’s preferences 132 ppp

[0, 1, 1] (A)

[0, 1, 0] (T) xxx

spspspV

332211]0,1,0[

spspspV

332211]1,1,0[

expected values of possible next election states:

],,[ 321 ssss

321 sss ],,[ 321 pppp

spspsp

332211

so T is better than A only when:

or, equivalently:

],,[ 321 ssss

321 sss ],,[ 321 pppp

spspsp

332211

so T is better than A only when:

or, equivalently:

Intuitively, T does better than A only when:

• s1 and s2 are relatively close

• x is relatively small

• p3 is relatively close to p1 compared to p2

],,[ 321 ssss

321 sss ],,[ 321 pppp

Corollaries:– When x is taken to infinity and , strategy A

dominates strategy T– When

, strategy A dominates strategy T

121 ss

32T is better than A only when:

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.

Further result for strategy A

More generally, it is true that if– the election state is free of ties and near-ties:

– and the focal voter’s cardinal preferences are tie-free:

when– 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)

121 321 kssss k

ji pp ji

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

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.

k > 1 m = 1 winner

Electing a committee from approval ballots

11110 00011

0000110111

•What’s the best committee of size m = 2?

approves ofcandidates

4 and 5k = 5 candidates

n = 6 ballots

Sum of Hamming distances

11110 00011

0000110111

01111 110004 5

4 3 sum = 22

m = 2 winners

Fixed-size minisum

11110 00011

0000110111

01111 00011

•Minisum elects winner set with smallest sumscore•Easy to compute (pick candidates with most approvals)

2 1 sum = 10

m = 2 winners

Maximum Hamming distance

11110 00011

0000110111

01111 000112 1

2 1 sum = 10max = 4

m = 2 winners

Fixed-size minimax

•Minimax elects winner set with smallest maxscore•Harder to compute?

11110 00011

0000110111

01111 001102 1

2 3 sum = 12max = 3

m = 2 winners

[Brams, Kilgour & Sanver ’04]

Complexity

Endogenous minimax

= EM = BSM(0, k)

Bounded-size minimax

= BSM(m1, m2)

Fixed-size minimax

= FSM(m) = BSM(m, m)

NP-hard

[Frances & Litman ’97]

NP-hard

(generalization of EM)

Complexity

Endogenous minimax

= EM = BSM(0, k)

= BSM(m1, m2)

Fixed-size minimax

NP-hard

[Frances & Litman ’97]

NP-hard

(generalization of EM)

NP-hard

(proof in written proposal)

Approximability

Endogenous minimax

= EM = BSM(0, k)

= BSM(m1, m2)

Fixed-size minimax

has a PTAS*

[Li, Ma & Wang ’99]

no known PTAS no known PTAS

* Polynomial-Time Approximation Scheme: algorithm with approx. ratio 1 + ε that runs in time polynomial in the input and exponential in 1/ε

Approximability

Endogenous minimax

= EM = BSM(0, k)

= BSM(m1, m2)

Fixed-size minimax

has a PTAS*

[Li, Ma & Wang ’99]

no known PTAS;

has a 3-approx.

no known PTAS;

has a 3-approx.

* Polynomial-Time Approximation Scheme: algorithm with approx. ratio 1 + ε that runs in time polynomial in the input and exponential in 1/ε

Approximating FSM

m = 2 winners

choosea ballot

arbitrarily

Approximating FSM

0010100111coerce to

size m

m = 2 winners

choosea ballot

arbitrarily

outcome =m-completed ballot

Approximation ratio ≤ 3

≤ OPT

optimalFSM set

OPT = optimal maxscore

00110 00111

≤ OPT ≤ OPT

optimalFSM set

chosenballot

OPT = optimal maxscore

00110 00111 00011

≤ OPT ≤ OPT ≤ OPT

≤ 3·OPT

optimalFSM set

chosenballot

m-completedballot

OPT = optimal maxscore (by triangle inequality)

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

Local search approach for FSM

1. Start with some c {0,1}k of weight m

010014

2. In c, swap up to r 0-bits with 1-bits in such a way that minimizes the maxscore of the result

11000 10001

01010 00011

001014

010104

3. Repeat step 2 until maxscore(c) is unchanged k times

4. Take c as the solution

11000 10010

01001 00011

001104

3. Repeat step 2 until maxscore(c) is unchanged k times

4. Take c as the solution

001103

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– highest average ratio < 1.04

• 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

(maxscore of solution found)(maxscore of exact solution)

Manipulating FSM

00110 00011

0000110111

01111 00011

•Voters are sincere

•Another optimal solution: 00101

max = 2

m = 2 winners

Manipulating FSM

11110 00011

0000110111

01111 00110

•A voter manipulates and realizes ideal outcome

•But our 3-approximation for FSM is nonmanipulable

max = 3

m = 2 winners

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

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

Completed: comparison of A and T in 3-alt. case, domination of A as

To do: comparisons of other pairs, analysis using branching-probabilities metric

Fixed-size minimax

Completed: NP-hardness proof, 3-approximation, heuristic evaluation, manipulability analysis

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?

What happens at equilibrium?

• The optimal strategy recommends that no voter change

• So• And

– equivalently,

• Therefore any average at equilibrium must satisfy two equations:– (A)– (B)

1)( ii vrvi

ii rvvi 0)(0)( ii vrvi

irvinv : nvrvi i :

Proof: Only one equilibrium average

irinA :)( nriB i :)(

212211 )()()()( BABA

• Theorem:

• Proof considers two symmetric cases:– assume– assume

• Each leads to a contradiction

21 case 1:

ii rri 12)( ii riri 12 :: ii riri 12 ::

irin 22 : nri i 11:

nririn ii 1122 :: nn 12

12 21 , contradicting

)( 2A)( 1B

21 Case 1 shows that

Case 2 is symmetrical and shows that 12

21 Therefore

Therefore, given , the average at equilibrium is uniquer

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

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)

Average approx. ratios found

radius = 1 radius = 2fixed-size minimax

1.0012 1.0000

3-approx. 1.0017 1.0000

random set

1.0057 1.0000

highest-maxscore

1.0059 1.0000

performance on GTS ’03 election data

k = 24 candidates, m = 12 winners, n = 161 ballots

Largest approx. ratios found

radius = 1 radius = 2fixed-size minimax

1.0714 1.0000

3-approx. 1.0714 1.0000

random set

1.0714 1.0000

highest-maxscore

1.0714 1.0000

performance on GTS ’03 election data

k = 24 candidates, m = 12 winners, n = 161 ballots

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)

computational aspects of approval voting and declared-strategy voting rob legrand washington...

set of ballots bu

distinguished member

kind of manipulation

borda conitzer sandholm

tweaks conitzer sandholm

tovey trick

hare bartholdi orlin

voterscbaplurality votingb

Documents

legrand part1

formations 2019 - legrand · hometouch. 3 legrand group...

cariva-legrand · title: cariva-legrand author: chorus -...

benjamin legrand - theatro-france.fr · benjamin legrand...

legrand magyarorszÁg

legrand elektro

valena legrand

legrand cabofil

legrand celian

legrand protecciones

price legrand electrica - instal & energy legrand...

legrand 2011

legrand 2013

catalog legrand

pial legrand

télecommandse legrand

celiane legrand

transientes legrand

emergencias legrand

legrand products, home & industrial products, legrand wires