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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 med (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 med (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 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



                                                                     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




RMSE a , 0.4820




                                a
                   minimum at   a  0.3647        48
      Evaluating AAR DSV systems: hill-climbing
                         a  0.3647




RMSE 0.3647 ,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                         x
                                                               x
                                   
     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   vn

                                                      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 
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
                                               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  r j
• 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 vi
                                
    will not change f aveq (v , m, M ) .


                                                             86
            AAR DSV is immune to strategy
               
• If   f aveq (v , m, M )  vi  ri,       
   – increasing vi will not change f aveq (v , m, M ).
                                              
   – decreasing vi will not increase f aveq (v , m, M ) .
            
• If   f aveq (v , m, M )  vi  ri,         
   – increasing vi will not decrease f aveq (v , m, M ) .
                                           
   – decreasing vi will not change f aveq (v , m, M ) .

                 (complete proof in dissertation)

• So voting sincerely ( vi  ri ) is guaranteed to
  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 med 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)
                    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




                                                                    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 s1  s2  1, strategy A
     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 p i 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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