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Computational Aspects of Approval Voting

and Declared-Strategy Voting



Dissertation defense

17 April 2008





Rob LeGrand

Washington University in St. Louis

Computer Science and Engineering

legrand@cse.wustl.edu







Ron Cytron Robert Pless

Steven Brams Itai Sened

Jeremy Buhler Aaron Stump

Themes of research



• Approval voting systems

• Susceptibility to insincere strategy

– encouraging sincere ballots

• Evaluating effectiveness of various strategies

• Internalizing insincerity

– separating strategy from indication of preferences

• Complex voting protocols

– complexity of finding most effective ballot

– complexity of calculating the outcome





2

What is “manipulation”?



• Broadly, effective influence on election outcome

• Election officials can . . .

– exclude/include alternatives [Nurmi ’99]

– exclude/include voters [Bartholdi, Tovey & Trick ’92]

– choose election protocol [Saari ’01]

• Alternatives may be able to . . .

– drop out to avoid a vote-splitting effect

• Voters can . . .

– find the ballot that is likeliest to optimize the outcome

• This last sense is what we mean

3

Let’s vote!



45 voters 35 voters 20 voters



A B C (1st)

sincere

preferences C C B (2nd)



B A A (3rd)









4

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

5

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

6

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

7

Manipulation decision problem



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

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?



• My generalization of problems from the literature:

[Bartholdi, Tovey & Trick ’89] [Conitzer & Sandholm ’02]

[Conitzer & Sandholm ’03]



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?



• 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

10

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

11

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 (single-winner STV) [Bartholdi & Orlin ’91]

– Borda [Conitzer & Sandholm ’02]

12

What can we do to make manipulation hard?



• 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









13

Declared-Strategy Voting

[Cranor & Cytron ’96]







cardinal rational

preferences strategizer



ballot



election

outcome

state









14

Declared-Strategy Voting

[Cranor & Cytron ’96]



sincerity strategy



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 sincere preferences

15

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



16

Can DSV be hard to manipulate?



DSV can be made to be NP-hard to manipulate in

the EPWCB sense. [LeGrand ’08]



Proof by reduction:

• Simulate Hare by using particular declared strategy in DSV

• Hare is NP-hard to manipulate [Bartholdi & Orlin ’91]

• If this DSV system were easy to manipulate, then Hare

would be

•  DSV can be made NP-hard to manipulate

So why use “tweaks”? (DSV is better!)





17

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

18

Approve both!



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

19

Approval voting

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





• Allows approval of any subset of alternatives

• Single alternative with most votes wins

• Used historically [Poundstone ’08]

– Republic of Venice 1268-1789

– Election of popes 1294-1621

• Used today [Brams ’08]

– Election of UN secretary-general

– Several academic societies, including:

• Mathematical Society of America

• American Statistical Association

20

Strands of 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

Strands of 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]

22

Strands of 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]

23

Approval ratings









24

Approval ratings









• Aggregating film reviewers’ ratings

– Rotten Tomatoes: approve (100%) or disapprove (0%)

– Metacritic.com: ratings between 0 and 100

– Both report average for each film

– Reviewers rate independently

25

Approval ratings









• Online communities

– Amazon: users rate products and product reviews

– eBay: buyers and sellers rate each other

– Hotornot.com: users rate other users’ photos

– Users can see other ratings when rating

• Can these “voters” benefit from rating insincerely?

26

Approval ratings









27

Average of ratings





r  0.4, 0.7, 0.8, 0.8, 0.9



v  0.4, 0.7, 0.8, 0.8, 0.9



outcome: f avg (v )  0.72



0.72

0 1









data from Metacritic.com: Videodrome (1983)

28

Average of ratings





r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0.7, 0.8, 0.8, 0.9



outcome: f avg (v )  0.64



0.64

0 1









Videodrome (1983)

29

Another approach: Median





r  0.4, 0.7, 0.8, 0.8, 0.9



v  0.4, 0.7, 0.8, 0.8, 0.9



outcome: f m ed (v )  0.8



0.8

0 1









Videodrome (1983)

30

Another approach: Median





r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0.7, 0.8, 0.8, 0.9



outcome: f m ed (v )  0.8



0.8

0 1









Videodrome (1983)

31

Another approach: Median



• Immune to insincerity [LeGrand ’08]

– voter i cannot obtain a better result by voting vi  ri 



– if f m ed (v )  vi , increasing vi will not change f m ed (v )

 

– if f m ed (v )  vi , decreasing vi will not change f m ed (v )





• Allows tyranny by a majority



– v  0, 0, 0,1,1,1,1



m ed (v )  1

– f

– no concession to the 0-voters







32

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?

• And will an equilibrium always be found?

33

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0.4, 0.7, 0.8, 0.8, 0.9



0.72

0 1









Videodrome (1983)

34

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0, 0, 0, 0



0

1









35

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0, 0, 0,1



0 .2

0 1









36

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0, 0,1,1



0 .4

0 1









37

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0,1,1,1



0 .6

0 1









38

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0.5,1,1,1

equilibrium!



0 .7

0 1









• Is this algorithm is guaranteed to find an equilibrium?



39

Equilibrium-finding algorithm



r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0.5,1,1,1

equilibrium!



0 .7

0 1









• Is this algorithm is guaranteed to find an equilibrium?

• Yes! [LeGrand ’08]

40

Expanding range of allowed votes



r  0.4, 0.7, 0.8, 0.8, 0.9



v  1, 1, 2, 2, 2



0.8

1 2









• These results generalize to any range [LeGrand ’08]



41

Multiple equilibria can exist



r  0.4, 0.7, 0.7, 0.8, 0.9



v  0, 0.5,1,1,1



v  0, 0.6, 0.9,1,1



v  0, 0.75, 0.75,1,1

outcome in each case:



f avg (v )  0.7

• Will multiple equilibria will always have the same average?



42

Multiple equilibria can exist



r  0.4, 0.7, 0.7, 0.8, 0.9



v  0, 0.5,1,1,1



v  0, 0.6, 0.9,1,1



v  0, 0.75, 0.75,1,1

outcome in each case:



f avg (v )  0.7

• Will multiple equilibria will always have the same average?

• Yes! [LeGrand ’08]

43

Average-Approval-Rating DSV





r  0.4, 0.7, 0.8, 0.8, 0.9



v  0.4, 0.7, 0.8, 0.8, 0.9



outcome: f aveq (v , 0,1)  0.7



0 .7

0 1









Videodrome (1983)

44

Average-Approval-Rating DSV





r  0.4, 0.7, 0.8, 0.8, 0.9



v  0, 0.7, 0.8, 0.8, 0.9



outcome: f aveq (v , 0,1)  0.7



0 .7

0 1









• AAR DSV is immune to insincerity in general [LeGrand ’08]

45

Evaluating AAR DSV systems



• Expanded vote range gives wide range of AAR



DSV systems:  a ,b (v ) 0  a  1 0  b  1

• If we could assume sincerity, we’d use Average

• Find AAR DSV system that comes closest

• Real film-rating data from Metacritic.com

– mined Thursday 3 April 2008

– 4581 films with 3 to 44 reviewers per film

– measure root mean squared error







46

Evaluating AAR DSV systems

b  0.5









RMSEa,0.5









a

minimum at a  0.3240 47

Evaluating AAR DSV systems: hill-climbing

b  0.4820









RMSEa,0.4820









a

minimum at a  0.3647 48

Evaluating AAR DSV systems: hill-climbing

a  0.3647









RMSE0.3647,b









b

minimum at b  0.4820 49

Evaluating AAR DSV systems











0.3647,0.4820(v )











f avg (v )

50

AAR DSV: Future work



• Website: ratingsbyrob.com

– Users can rate movies, books, each other, etc.

– They can see current ratings without being tempted to

rate insincerely

• Find more strategy-immune rating systems

• Richer outcome spaces

– Hypercube: like rating several films at once

– Simplex: dividing a limited resource among several uses

– How assumptions about preferences are generalized is

important





51

Strands of 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]

52

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





53

DSV-style approval strategies



s  [30, 25,15,10]



 p  [0, 1, 0.8, 0.3]

• Strategy Z: b  [0, 1, 1, 0]

– Approve alternatives with higher-than-average cardinal

preference (zero-information strategy) [Merrill ’88]









54

DSV-style approval strategies



s  [30, 25,15,10]



 p  [0, 1, 0.8, 0.3]

• Strategy Z: b  [0, 1, 1, 0]



• Strategy T: b  [0, 1, 0, 0]

– Approve favorite of top two vote-getters, plus all liked

more [Ossipoff ’02, Poundstone ’08]

– Simplest generalization of plurality DSV strategy

[Cranor & Cytron ’96]









55

DSV-style approval strategies



s  [30, 25,15,10]



 p  [0, 1, 0.8, 0.3]

• Strategy Z: b  [0, 1, 1, 0]



• Strategy T: b  [0, 1, 0, 0]



• Strategy J: b  [0, 1, 1, 0]

– Use strategy Z if it distinguishes between top two vote-

getters; otherwise use strategy T [Brams & Fishburn ’83]









56

DSV-style approval strategies



s  [30, 25,15,10]



 p  [0, 1, 0.8, 0.3]

• Strategy Z: b  [0, 1, 1, 0]



• Strategy T: b  [0, 1, 0, 0]



• Strategy J: b  [0, 1, 1, 0]



• Strategy A: b  [0, 1, 1, 1]

– Approve all preferred to top vote-getter, plus top vote-

getter if preferred to second-highest vote-getter

[LeGrand ’02]

. . . but how to evaluate these strategies?



57

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?









58

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

Wi  k

i





s

j 1

x

j









59

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)



60

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



When p2  p3  p1 , A is better than T if and only if:

p1s1x  p2 s2  1  p3 s3  1 p1s1x  p2 s2  1  p3 s3

x x xx



s1  s2  1  s3  1 s1x  s2  1  s3

x x x x x







or, equivalently:

x

p2  p3  s1  Intuitively, A always does better than T when:

 s 1

  • s1 is much larger than s2,

p3  p1  2  • x is large, or

• p3 is relatively close to p2 compared to p1



61

Strategy comparison using the Merrill metric



• Also compared other strategy pairs [LeGrand ’08]

• As x goes to infinity (3 alternatives):

– Strategy A dominates strategy T

– Strategy A dominates strategy J

– Strategy A dominates strategy Z

– Neither strategy T nor strategy J dominates the other

• As x goes to infinity (4 alternatives):

– Strategy A dominates strategy T









62

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 approval

strategies according to the Merrill metric [LeGrand ’08]









63

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









64

Branching-probabilities metric: strategy A



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 number of future ballots is taken to infinity

then strategy A dominates all other approval

strategies according to the branching-probabilities

metric [LeGrand ’08]





65

Approval DSV strategies: Future work



• Consider different strategy-evaluation metrics

• Study strategy-A equilibria

– How “good” are the outcomes?

– How often are strong Nash equilibria found?

• How strategy-vulnerable is Approval DSV with

strategy A?

– How often will submitting insincere preferences benefit a

voter?









66

Strands of 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]

67

Electing a committee from approval ballots



approves of

k = 5 alternatives 11110 00011 alternatives

4 and 5

n = 6 ballots





01111 00111









10111 00001







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

68

Sum of Hamming distances





m = 2 winners 11110 00011



2 4



4 5

01111 11000 00111



4 3 sum = 22





10111 00001







•What if we elect alternatives 1 and 2?

69

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 HD sum

•Easy to compute (pick alternatives with most approvals)

70

Maximum Hamming distance





m = 2 winners 11110 00011



4 0



2 1

01111 00011 00111



2 1 sum = 10

max = 4

10111 00001







•One voter is quite unhappy with minisum outcome

71

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 maximum HD

•Harder to compute?

72

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)









73

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) [LeGrand ’04]









74

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

75

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]

[LeGrand, Markakis & [LeGrand, Markakis &

Mehta ’06] Mehta ’06]



* Polynomial-Time Approximation Scheme: algorithm

with approx. ratio 1 + ε that runs in time polynomial in

the input and exponential in 1/ε

76

Susceptibility to insincerity







Endogenous minimax Bounded-size minimax Fixed-size minimax

= EM = BSM(0, k) = BSM(m1, m2) = FSM(m) = BSM(m, m)







insincere voters insincere voters insincere voters

can benefit can benefit can benefit



[LeGrand, Markakis & [LeGrand, Markakis & [LeGrand, Markakis &

Mehta ’06] Mehta ’06] Mehta ’06]



But our 3-approximation for FSM is

immune to insincere strategy!

77

Fin



Thanks to

– my advisor, Ron Cytron

– Steven Brams

– members of my committee

– co-authors Vangelis Markakis and Aranyak Mehta

– Morgan Deters and the rest of the DOC Group







Questions?





78

Rational [m,M]-Average strategy



• Allow votes between m  0 and M  1

• 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 , m), M )



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



– f avg (v )  ri and vi  M



– f avg (v )  ri



– f avg (v )  ri and vi  m

79

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   v n



80

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



81

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 

2 n  i : 2  ri  A(2 )

i : 1  ri   1n B(1 )

2 n  i : 2  ri   i : 1  ri   1n

2n  1n

2  1 , contradicting 1  2

82

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









83

An equilibrium always exists?



• At equilibrium, v must satisfy

(i) vi  min(max( ri n   j i v j , m), M )



I proposed to prove that, given a vector r , at least

one equilibrium exists.



A particular algorithm will always find an equilibrium



for any r . . .







84

An equilibrium always exists!



Equilibrium-finding algorithm:



• sort r so that (i  j ) ri  rj

• for i = 1 up to n do

vi  min(max( ri n  k i vk  (n  i)m, m), M )



(full proof and more efficient algorithm in dissertation)



• Since an equilibrium always exists, average at



equilibrium is a function, f aveq (r , m, M ) .

 

• Applying f aveq to v instead of r gives a new

system, Average-Approval-Rating DSV.

85

Average-Approval-Rating DSV



• What if, under AAR DSV, voter i could gain an

outcome closer to ideal by voting insincerely

( vi  ri )?



I proposed to prove that Average-Approval-Rating

DSV is immune to strategy by insincere voters.



• Intuitively, if f aveq (v , m, M )  vi, increasing v i



will not change f aveq (v , m, M ) .





86

AAR DSV is immune to strategy



• If f aveq (v , m, M )  vi  ri, 

– increasing v i will not change f aveq (v , m, M ).



– decreasing v i will not increase f aveq (v , m, M ) .



• If f aveq (v , m, M )  vi  ri, 

– increasing v i will not decrease f aveq (v , m, M ) .



– decreasing v i will not change f aveq (v , m, M ) .



(complete proof in dissertation)



• So voting sincerely ( v  r ) is guaranteed to

i i

optimize the outcome from voter i’s point of view



87

Parameterizing AAR DSV



• [m,M]-AAR DSV can be parameterized nicely using

a and b, where 0  a  1 and 0  b  1:

1 m

a b

M m 1 M  m

b 1 b

m b M b

a a

  b 1 b 

 a ,b (v )  lim f aveq v , b  , b  

x a  x x 

88

Parameterizing AAR DSV



• For example:

 

1,b (v )  f aveq (v , 0,1)

 

 1 1 (v )  f aveq v ,  1, 2

,

3 2  

 1 1 (v )  f aveq v ,  10,11

,

21 2

 

 1 (v )  f m ed v 

0,

2  

0,0 (v )  maxv 

 

0,1 (v )  min v  89

Evaluating AAR DSV systems



• Real film-rating data from Metacritic.com

– mined Thursday 3 April 2008

– 4581 films with 3 to 44 reviewers per film



0  a 1 0  b 1

SEa,b v    a,b v   f avg v 

   2



 

 v  SEa,b v 

RMSEa ,b V   vV





v 

vV 90

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









91

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]



92

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)

s1x  s2  1  s3  1

x x







p1s1x  p2 s2  1  p3 s3x

x

V[ 0,1,0 ]  [0, 1, 0] (T)

s1x  s2  1  s3x

x









93

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 





94

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 s  s  1 strategy A

,

1 2

dominates strategy T

– When p1  p2

p3  , strategy A dominates strategy T

2

95

Further result for strategy A





V[ 0, 0,...0]   p1 , p2 , pk 

s1 , s2 , sk 

s1  s2    sk

• just a weighted average of pi values

• assume p1  p2

• as x   , V[ 0,0,...0]  p1 from below

• so maximized when weights of those pi  p1 are

maximized, which is done by approving only alternatives i

where pi  p1

• p2  p1 case is similar: approve i where pi  p1

• only strategy A always does this



96

Approximating FSM





11110 m = 2 winners



00011



00111

00111

00001 choose

a ballot

10111

arbitrarily

01111









97

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







98

Approximation ratio ≤ 3



optimal

11110

2 FSM set

00011 2



00111 1

00110

3

00001

2

10111

2

01111

≤ OPT





OPT = optimal maxscore

99

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

100

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

101

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









102

Local search approach for FSM



1. Start with some c  {0,1}k

of weight m





01001

4









103

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









104

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









105

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









106

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







107

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







108

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





109

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 dissertation)



110

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









111

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



112

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



113

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)









114

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 115

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

116

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









117

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, strategy-immunity of

AAR DSV, evaluation of AAR DSV systems



DSV-style Completed: Merrill-metric comparison of A and T in 3-

approval alt. case, 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







118


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