1 Introduction
Computational social choice studies computational aspects of voting. For example, how does a coalition of agents compute a manipulation? Can we compile these votes into a more compact form? How do we decide if we have elicited enough votes from the agents to be able to declare the result? Whilst there has been a very active research community studying these sort of questions for well known voting rules like plurality and Borda, there are other less well known rules that might deserve attention. In particular, we put forward two historical voting rules due to Nanson and Baldwin which are related to Borda voting.
There are several reasons to consider these two rules. Firstly, they have features that might appeal to the two opposing camps that support Borda and Condorcet. In particular, both rules are Condorcet consistent as they elect the candidate who beats all others in pairwise elections. Secondly, both rules are elimination style procedures where candidates are successively removed. Other elimination procedures like STV and plurality with runoff are computationally hard to manipulate (in the case of STV, with or without weights on the votes, whilst in the case of plurality with runoff, only in the case of weighted votes). We might therefore expect Nanson’s and Baldwin’s rules to be computationally hard to manipulate. Thirdly, statistical analysis suggest that, whilst the Borda rule is vulnerable to manipulation [Chamberlin1985], Nanson’s rule is particularly resistant [Favardin & Lepelley2006]. We might expect Baldwin to be similarly resistant. Finally, the two rules have been used in real elections in the Universitiy of Melbourne (between 1926 and 1982), the University of Adelaide (since 1968), and the State of Michigan (in the 1920s). It is perhaps therefore somewhat surprising that neither rule has received much attention till now in the computational social choice literature.
2 Preliminaries
Let be the set of candidates (or alternatives). A linear order on is a transitive, antisymmetric, and total relation on . The set of all linear orders on is denoted by . An voter profile on consists of linear orders on . That is, , where for every , . The set of all profiles is denoted by . We let denote the number of candidates. A (deterministic) voting rule is a function that maps any profile on to a unique winning candidate, that is, . In this paper, if not mentioned otherwise, ties are broken in the fixed order .
(Positional) scoring rules are commonly used voting rules. Each positional scoring rule is identified by a scoring vector of integers, for any vote and any candidate , let , where is the rank of in . For any profile , let . The rule selects such that the total score is maximized. We assume scores are integers and decreasing. Borda
is the positional scoring rule that corresponds to the scoring vector
. We write for the Borda score given to candidate from the profile of votes , and where is obvious from the context. When voters are weighted (that is, each voter is associated with a positive real number as the weight), a positional scoring rule selects the candidate that maximizes the weighted total score.The unweighted (coalitional) manipulation problem is defined as follows. An instance is a tuple , where is a voting rule, is the nonmanipulators’ profile, is the candidate preferred by the manipulators, and is the set of manipulators. We are asked whether there exists a profile for the manipulators such that . The weighted (coalitional) manipulation is defined similarly, where the weights of the voters (both nonmanipulators and manipulators) are also given as inputs. As is common in the literature, we break ties in favour of the coalition of the manipulators where appropriate.
3 Nanson’s and Baldwin’s Rules
The Borda rule has several good properties. For instance, it is monotonic as increasing the score for a candidate only helps them win. Also it never elects the Condorcet loser (a candidate that loses to all others in a majority of head to head elections). However, it may not elect the Condorcet winner (a candidate that beats all others in a majority of head to head elections). Nanson’s and Baldwin’s rules, by comparison, always elect the Condorcet winner when it exists.
Nanson’s and Baldwin’s rules are derived from the Borda rule. Nanson’s rule eliminates all those candidates with less than the average Borda score [Nanson1882]. The rule is then repeated with the reduced set of candidates until there is a single candidate left. A closely related voting rule proposed by Baldwin successively eliminates the candidate with the lowest Borda score^{1}^{1}1If multiple candidates have the lowest score, then we use a tiebreaking mechanism to eliminate one of them. until one candidate remains [Baldwin1926]. The two rules are closely related, and indeed are sometimes confused. One of the most appealing properties of Nanson’s and Baldwin’s rules is that they are Condorcet consistent, i.e. they elect the Condorcet winner. This follows from the fact that the Borda score of the Condorcet winner is never below the average Borda score. Both rules possess several other desirable properties including the majority criterion and the Condorcet loser criterion. There are also properties which distinguish them apart. For instance, Nanson’s rule satisfies reversal symmetry (i.e. if there is a unique winner and voters reverse their vote then the winner changes) but Baldwin’s rule does not.
4 Unweighted Manipulation
We start by considering the computational complexity of manipulating both these rules with unweighted votes. We prove that the coalitional manipulation problem is NPcomplete for both rules even with a single manipulator. Computational intractability with a single manipulator is known only for a small number of other voting rules including the second order Copeland rule [Bartholdi, Tovey, & Trick1989], STV [Bartholdi and Orlin1991] and ranked pairs [Xia et al.2009]. In contrast, when there are two or more manipulators, unweighted coalitional manipulation is hard for some other common voting rules [Faliszewski, Hemaspaandra, and Schnoor2008, Faliszewski, Hemaspaandra, and Schnoor2010, Xia, Conitzer, and Procaccia2010, Davies et al.2011, Betzler, Niedermeier, and Woeginger2011]. Our results therefore significantly increase the size of the set of voting rules used in practice that are known to be NPhard to manipulate with a single manipulator. This also contrasts to Borda where computing a manipulation with a single manipulator is polynomial [Bartholdi, Tovey, & Trick1989]. Adding elimination rounds to Borda to get Nanson’s or Baldwin’s rules increases the computational complexity of computing a manipulation with one manipulator from polynomial to NPhard.
Our results are proved by reductions from the exact 3cover (x3c) problem. An x3c instance contains two sets: and , where and for all , and . We are asked whether there exists a subset of such that each element in is in exactly one of the 3sets in .
Theorem 1.
With unweighted votes, the coalitional manipulation problem under Baldwin’s rule is NPcomplete even when there is only one manipulator.
Proof: We sketch a reduction from x3c. Given an x3c instance , we let the set of candidates be , where is the candidate that the manipulator wants to make the winner, , and and are additional candidates. Members of correspond to the 3sets in . Let .
The profile contains two parts: , which is used to control the changes in the score differences between candidates, after a set of candidates are removed, and , which is used to balance the score differences between the candidates. We define the votes where Others is a total order in which the candidates in are in a predefined lexicographic order, and is the reverse.
We make the following observations on . For any set of candidates and any pair of candidates ,
Here is the pair of votes obtained from by removing all candidates in . In words, the formula states that after is removed, the score difference between and is increased by if and only if and is removed; it is decreased by if and only if and is removed; for any other cases, the score difference does not change. Moreover, for any , and .
We next show how to use to construct the first part of the profile . Let , that is, . is composed of the following votes: (1) for each and each , there are copies of ; (2) for each , there are copies of ; (3) there are copies of . It is not hard to verify that , and for any , . is composed of the following votes: (1) for each , there are copies of ; (2) for each , there are copies of ; (3) there are copies of .
Let . We make the following observations on the Borda scores of the candidates in .
For any , ;
for any , ;
.
Suppose the x3c instance has a solution, denoted by (after reordering the sets in ) . Then, we let the manipulator vote for:
In the first rounds, all candidates in and drop out. Then drops out. In the following rounds the candidates in drop out. Finally, loses to in their pairwise election, which means that is the winner.
Suppose the manipulator can cast a vote to make the winner. We first note that must be eliminated in the final round since its score is higher than in all previous rounds. In the round when is eliminated, the score of should be no more than the score of . We note that and the score difference can only be reduced by the manipulator ranking below , and by eliminating before . However, by ranking below , the score difference is reduced by no more than . Therefore, before drops out, all candidates in must have already dropped out. We note that for any , . Therefore, for each , there exists with who is removed before . For any such , none of the candidates in can drop out before (otherwise the score of cannot be less than before drops out), and in the next three rounds the candidates in drop out. It follows that the set of candidates in that drop out before any candidate in corresponds to an exact cover of .
Theorem 2.
With unweighted votes, the coalitional manipulation problem under Nanson’s rule is NPcomplete even when there is only one manipulator.
The proof uses the same gadget that is used in the proof of Theorem 1. Due to the space constraints, the proof can be found in an online technical report.
Weighted Manipulation
If the number of candidates is bounded, then manipulation is NPhard to compute when votes are weighted. Baldwin’s rule appears more computationally difficult than Nanson’s rule. Coleman07:Complexity Coleman07:Complexity prove that Baldwin’s requires only 3 candidates to be NPhard, whilst we prove here that Nanson’s rule is polynomial to manipulate with 3 candidates and requires at least 4 candidates to be NPhard. It follows that computing a manipulation is NPhard for both rules when votes are unweighted, the number of candidates is small and there is uncertainty about how agents have voted in the form of a probability distribution
[Conitzer, Sandholm, and Lang2007]. Note that the coalition manipulation problem for Borda with weighted votes is NPhard for 3 or more candidates [Conitzer, Sandholm, and Lang2007]. Thus, somewhat surprisingly, adding an elimination round to Borda, which gives us Nanson’s rule, decreases the computational complexity of computing a manipulation with 3 manipulators from NPhard to polynomial.Theorem 3.
With Nanson’s rule and weighted votes, the coalition manipulation problem is NPcomplete for just 4 candidates.
Proof: The proof is by a reduction from partition, where we are given a group of integers with sum , and we are asked whether there is way to partition the group into two groups, the elements in each of which sum to . For any partition instance, we construct a coalition manipulation problem with 4 candidates (, , and ) where is again the candidate that the manipulators wish to win. We suppose the nonmanipulators have voted as follows: for each of , , and , for and , and 1 each for , , and . The total scores from nonmanipulators are as follows: , and . For each integer , we have a member of the manipulating coalition with weight .
Now, suppose there is a solution to the partition instance. Let the manipulators corresponding to the integers in one half of the partition vote , and let the others vote . All scores are now (which is also the average). By the tiebreaking rule, wins in the first round. Thus the manipulators can make win if a perfect partition exists.
Conversely, suppose there is a successful manipulation. Clearly, cannot be eliminated in the first round. To ensure this, all manipulators must put in first place. Next, we show that if is not a joint winner of the first round, cannot win overall. We consider all possible sets of candidates that could be eliminated in the first round. There are 6 cases. In the first case, only is eliminated in the first round. The scores from nonmanipulators in the second round are as follows: , and . The average score is . Even with the maximum possible score from the manipulators, is eliminated. This contradicts the assumption that wins. In the second case, only is eliminated in the first round. As and are not eliminated in the first round, the manipulators have to cast votes that put in first place and in second place. With such manipulating votes, the scores in the second round are: , and . The average score is . Hence, is eliminated. In the next round, is eliminated as , and the average score is . This contradicts the assumption that wins. In the third case, only is eliminated in the first round. This case is symmetric to the second case. In the fourth case, and are eliminated in the first round. The case when and are eliminated is symmetric. In the second round, the scores from nonmanipulators are and . The score from the manipulators cannot prevent being eliminated. This contradicts the assumption that wins. In the fifth case, and are eliminated in the first round. However, in the first round, the score and receive from the nonmanipulators is . One of them will get at least points from manipulators. This will give them greater than the average score of . Hence, at least one of them is not eliminated. In the sixth and final case, , and are all eliminated in the first round. This case is again impossible by the same argument as the last case.
The only way for to win is to have a tie with all candidates in the first round. As we observed above, the manipulators have to put in first place, and in second place. In turn, both and have to get exactly points from the manipulators. Hence, there exists a solution to the partition instance.
Clearly, it is polynomial to compute a manipulation of Baldwin’s rule with 2 candidates (since this case degenerates to majority voting). With Nanson’s rule, on the other hand, it is polynomial with up to 3 candidates.
Theorem 4.
With Nanson’s rule and weighted votes, the coalition manipulation problem is polynomial for up to 3 candidates.
Proof: Consider an election with 3 candidates ( and ) in which the manipulators want to win. We prove that the optimal strategy is for the manipulators either all to vote or all to vote . If does not win using one of these two votes, then cannot win. Therefore we simply try out the two votes and compute if wins in either case.
Suppose the manipulators can make win. We first note that there is no loss for them to raise to the first position, while keeping the other parts of their preferences the same. By doing so, the score of goes up and the scores of and go down. The only possible change in the elimination process is that now both and drop out in the first round, so that still wins.
Now, suppose that all manipulators rank in their top positions. Let denote the manipulators’ profile that makes win. Because Nanson’s rule never selects the Condorcet loser, cannot be beaten by both and in pairwise elections. Without loss of generality, suppose beats . We argue that if all manipulators vote , then still wins. For the sake of contradiction, suppose all manipulators vote but does not win. As the manipulators still rank in their top positions, the score of in the first round is the same as in . Therefore, must enter (and lose) the second round. Hence, only is eliminated in the first round, and in the second round beats . However, having the manipulators vote only lowers ’s score in the first round, compared to the case where they vote . Hence, when the manipulators vote , also enters the second round and then beats , which is a contradiction.
Therefore, if the manipulators can make win, then they can make win by all voting , or all voting .
5 Approximation Methods
One way to deal with computational intractability is to treat computing a manipulation as an approximation problem where we try to minimize the number of manipulators. We therefore considered five approximation methods. These are either derived from methods used with Borda or are specifically designed for the elimination style of Nanson’s and Baldwin’s rules.
 Reverse:

The desired candidate is put first, and the other candidates are reverse ordered by their current Borda score. We repeat this construction until the desired candidate wins. Reverse was used to manipulate the Borda rule in [Zuckerman, Procaccia, and Rosenschein2009].
 LargestFit:

This method was proposed for the Borda rule [Davies et al.2010]. Unlike Reverse
which constructs votes one by one, we construct votes in any order using a bin packing heuristic which puts the next largest Borda score into the “best” available vote. We start with a target number of manipulators. Simple counting arguments will lower bound this number, and we can increae it until we have a successful manipulation. We construct votes for the manipulators in which the desired candidate is in first place. We take the other Borda scores of the manipulators in decreasing order, and assign them to the candidate with the lowest current Borda score who has been assigned less than the required number of scores. A perfect matching algorithm then converts the sets of Borda scores for the candidates into a set of manipulating votes.
 AverageFit:

This method was also proposed for the Borda rule [Davies et al.2010]. We again have a target number of manipulators, and construct votes for the manipulators in which the desired candidate is in first place. We take the other Borda scores of the manipulators in decreasing order, and assign them to the candidate with the current lowest average Borda score who has less than the required number of scores. The intuition is that if every score was of average size, we would have a perfect fit. If more than one candidate has the same lowest average Borda score and can accommodate the next score, we tiebreak on the candidate with the fewest scores. Examples of LargestFit and AverageFit can be found in [Davies et al.2010].
 Eliminate:

We repeatedly construct votes in which the desired candidate is put in first place, and the other candidates in the reverse of the current elimination order. For instance, the first candidate eliminated is put in last place. For Nanson’s rule, we order candidates eliminated in the same round by their Borda score in that round.
 RevEliminate:

We repeatedly construct votes in which the desired candidate is put in first place, and the other candidates in the current elimination order. For instance, the first candidate eliminated is put in second place. For Nanson’s rule, we order candidates eliminated in the same round by the inverse of their Borda score in that round.
The intuition behind Eliminate is to move the desired candidate up the elimination order whilst keeping the rest of the order unchanged. With RevEliminate, the intuition is to move the desired candidate up the elimination order, and to assign the largest Borda scores to the least dangerous candidates. It is easy to show that all methods will eventually compute a manipulation of Nanson’s or Baldwin’s rule in which the desired candidate wins.
With Borda voting, good bounds are known on the quality of approximation that is achievable. In particular, [Zuckerman, Procaccia, and Rosenschein2009] proved that Reverse never requires more than one extra manipulator than optimal. Baldwin’s and Nanson’s rules appear more difficult to approximate within such bounds. We can give examples where all five methods compute a manipulation that use several more manipulators than is optimal. Indeed, even with a fixed number of candidates, Reverse can require an unbounded number of extra manipulators.
Theorem 5.
With Baldwin’s rule, there exists an election with 7 candidates and votes where Reverse computes a manipulation with at least more votes than is optimal.
Proof: (Sketch) Consider an election over , , , , , and where is the candidate that the manipulators wish to win. We define as the pair of votes: , where Others is some fixed ordering of the other candidates and is its reverse. The nonmanipulators cast the following votes: copies of , , , and . In addition, there are copies of the votes: and . If manipulators vote identically then wins. This provides an upper bound on the size of the optimal manipulation. After the nonmanipulators have voted, , and . Reverse will put in first place. We suppose is a multiple of 2, but more complex arguments can be given in other cases. After manipulating votes have been constructed, the scores of candidates to are level at and is leveled at . From then on, the manipulators put in first place and alternate the order of the other candidates. At least votes are therefore required for to move out of last place.
Asymptotically this result is as bad as we could expect. Any election can be manipulated with votes by simply reversing all previous votes, and this proof demonstrates that Reverse may use more votes than is optimal.
6 Experimental Results
Rules  Rev  LaFit  AvFit  Elim  RevElim 

Baldwin  74.4%  74.4%  75.8%  62.2%  75.2% 
Nanson  74.6%  76.0%  78.0%  65.4%  66.9% 
Borda  95.7%  98.8%  99.8%  95.7%  10.7% 
Rules  Rev  LaFit  AvFit  Elim  RevElim 

Baldwin  75.1%  75.4%  77.3%  68.9%  83.4% 
Nanson  78.1%  79.0%  79.8%  72.2%  79.4% 
Borda  96.1%  92.7%  99.9%  96.1%  4.4% 
To test the difficulty of computing manipulations in practice and the effectiveness of these approximation methods, we ran some experiments using a similar setup to [Walsh2010]. We generated either uniform random votes or votes drawn from a Polya Eggenberger urn model. In the urn model, votes are drawn from an urn at random, and are placed back into the urn along with other votes of the same type. This captures varying degrees of social homogeneity. We set so that there is a 50% chance that the second vote is the same as the first.
Our first set of experiments used 3000 elections with 5 candidates and 5 nonmanipulating voters. This is small enough to find the optimal number of manipulators using brute force search, and thus to determine how often a heuristic computes the optimal solution. We threw out the 20% or so of problems generated in which the chosen candidate has already won before the manipulators vote. Results are given in Tables 2–2. Heuristics that are very effective at finding an optimal manipulation with the Borda rule do not perform as well with Baldwin’s and Nanson’s rules. For example, AverageFit almost always finds an optimal manipulation of the Borda rule but can only find an optimal solution about 3/4 of the time with Baldwin’s or Nanson’s rules.
n  Rev  LaFit  AvgFit  Elim  RevElim 

4  2.25  2.25  2.25  2.44  2.21 
8  2.99  3.07  3.01  3.35  3.06 
16  4.31  4.41  4.40  4.79  4.67 
32  5.93  6.03  6.14  6.61  6.84 
64  8.56  8.65  8.84  9.54  11.02 
128  12.13  12.24  12.41  13.37  16.06 
n  Rev  LaFit  AvgFit  Elim  RevElim 

4  2.15  2.17  2.15  2.25  2.28 
8  2.91  2.96  2.84  3.05  3.21 
16  4.13  4.27  4.05  4.44  4.99 
32  5.80  5.88  5.81  6.18  7.46 
64  8.51  8.58  8.82  8.99  12.04 
128  12.07  12.09  13.00  12.60  17.90 
n  Rev  LaFit  AvgFit  Elim  RevElim 

4  3.26  3.23  3.24  3.35  3.14 
8  5.95  5.96  5.99  6.37  5.82 
16  11.64  11.66  11.87  12.74  11.52 
32  21.70  21.78  22.35  24.67  22.41 
64  43.09  43.37  44.24  49.07  45.70 
128  82.19  81.82  83.62  95.37  91.80 
n  Rev  LaFit  AvgFit  Elim  RevElim 

4  3.20  3.19  3.20  3.28  3.22 
8  5.93  5.98  5.95  6.13  6.09 
16  11.62  11.93  11.64  12.16  12.37 
32  22.36  22.78  22.53  24.00  24.39 
64  44.56  45.50  44.77  48.81  49.69 
128  87.18  87.55  86.76  97.02  99.43 
Our second set of experiments used larger problems. This amplifies the differences between the different approximation methods (but means we are unable to compute the optimal manipulation using brute force search). Problems have between and candidates, and the same number of votes as candidates. We tested 6000 instances, 1000 at each problem size. Tables 3–6 show the results for the average number of manipulators. The results show that overall Reverse works slightly better than LargestFit and AverageFit, which themselves outperform the other two methods especially for problems with large number of candidates. We observe a similar picture with Nanson’s rule. This contrasts with the Borda rule where LargestFit and AverageFit do much better than Reverse [Davies et al.2010]. In most cases AverageFit is less effective than LargestFit except urn elections with Nanson’s rule.
These experimental results suggest that Baldwin’s and Nanson’s rules are harder to manipulate in practice than Borda. Approximation methods that work well on the Borda rule are significantly less effective on these rules. Overall, Reverse, LargestFit and AverageFit appear to offer the best performance, though no heuristic dominates.
7 Other Related Work
weakestlink weakestlink prove that a class of voting rules which use repeated ballots and eliminate one candidate in each round are Condorcet consistent. They illustrate this class with the weakest link rule in which the candidate with the fewest ballots in each round is eliminated. geller geller has proposed a variant of single transferable vote where first place votes, candidates are successively eliminated based on their original Borda score. Unlike Nanson’s and Baldwin’s rules, this method does not recalculate the Borda score based on the new reduced set of candidates. For any Condorcet consistent rule (and thus for Nanson’s and Baldwin’s rule), bbhhaaai10 bbhhaaai10 showed that many types of control and manipulation are polynomial to compute when votes are single peaked.
8 Conclusions
With unweighted votes, we have proven that Nanson’s and Baldwin’s rules are NPhard to manipulate with one manipulator. This increases by two thirds the number of rules known to be NPhard to manipulate with just a single manipulator. With weighted votes, on the other hand, we have proven that Nanson’s rule is NPhard to manipulate with just a small number of candidates and a coalition of manipulators. We have also proposed a number of approximation methods for manipulating Nanson’s and Baldwin’s rules. Our experiments suggest that both rules are difficult to manipulate in practice. There are many other interesting open questions coming from these results. For example, are there other elimination style voting rules which are computationally difficult to manipulate? As a second example, with Nanson’s and Baldwin’s rule what is the computational complexity of other types of control like the addition/deletion of candidates, and the addition/deletion of voters? As a third example, we could add elimination rounds to other scoring rules. Do such rules have interesting computational properties?
Acknowledgments
Nina Narodytska is supported by the Asian Office of Aerospace Research and Development through grant AOARD104123. Toby Walsh is funded by the Australian Department of Broadband, Communications and the Digital Economy and the ARC. Lirong Xia acknowledges a James B. Duke Fellowship and Vincent Conitzer’s NSF CAREER 0953756 and IIS0812113, and an Alfred P. Sloan fellowship for support. We thank all AAAI11 reviewers for their helpful comments and suggestions.
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