proposal

Document Sample

```					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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