 # Combining Voting Rules Together

We propose a simple method for combining together voting rules that performs a run-off between the different winners of each voting rule. We prove that this combinator has several good properties. For instance, even if just one of the base voting rules has a desirable property like Condorcet consistency, the combination inherits this property. In addition, we prove that combining voting rules together in this way can make finding a manipulation more computationally difficult. Finally, we study the impact of this combinator on approximation methods that find close to optimal manipulations.

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## 1 Introduction

An attractive idea in the Zeitgeist of contemporary culture is “The Wisdom of Crowds” 

. This is the idea that, by bringing together diversity and independence of opinions, groups can be better at making decisions than the individuals that make up the group. For example, in 1907 Galton observed the wisdom of the crowd at guessing the weight of an ox in the West of England Fat Stock and Poultry Exhibition. The median of the 787 estimates was 1207 lb, within 1% of the correct weight of 1198 lb. We can view different voting rules as having different opinions on the “best” outcome to an election. We argue here that it may pay to combine these different opinions together. We provide both theoretical and experimental evidence for this thesis. On the theoretical side, we argue that a combination of voting rules can inherit a desirable property like Condorcet consistency when only one of the base voting rules is itself Condorcet consistent. We also prove that combining voting rules together can make strategic voting more computationally difficult. On the experimental side, we study the impact of combining voting rules on the performance of approximation methods for constructing manipulations.

### 1.1 Related Work

Different ways of combining voting rules to make manipulation computationally hard have been investigated recently. Conitzer and Sandholm  studied the impact on the computational complexity of manipulation of adding an initial round of the Cup rule to a voting rule. This initial round eliminates half the candidates and makes manipulation NP-hard to compute for several voting rule including plurality and Borda. Elkind and Lipmaa  extended this idea to a general technique for combining two voting rules. The first voting rule is run for some number of rounds to eliminate some of the candidates, before the second voting rule is applied to the candidates that remain. They proved that many such combinations of voting rules are NP-hard to manipulate. Note that theirs is a sequential combinator, in which the two rules are run in sequence, whilst ours (as we will see soon) is a parallel combinator, in which the two rules are run in parallel. More recently, Walsh and Xia  showed that using a lottery to eliminate some of the voters (instead of some of the candidates) is another mechanism to make manipulation intractable to compute.

## 2 Voting Rules

Voting is a general mechanism to combine together the preferences of agents. Many different voting rules have been proposed over the years, providing different opinions as to the “best” outcome of an election. We formalise voting as follows. A profile is a sequence of total orders over candidates. A voting rule is a function mapping a profile onto one candidate, the winner. Let be the number of voters preferring to in . Where is obvious from the context, we write . Let be 1 iff and 0 otherwise. We consider some of the most common voting rules.

Positional scoring rules: Given a scoring vector of weights, the th candidate in a vote scores , and the winner is the candidate with highest total score over all the votes. The plurality

rule has the weight vector

, the veto rule has the vector , the k-approval rule has the vector containing 1s, and the Borda rule has the vector .

Cup: The winner is the result of a series of pairwise majority elections between candidates. Given the agenda, a binary tree in which the roots are labelled with candidates, we label the parent of two nodes by the winner of the pairwise majority election between the two children. The winner is the label of the root.

Copeland: The candidate with the highest Copeland score wins. The Copeland score of candidate is . The Copeland winner is the candidate that wins the most pairwise elections.

Maximin: The maximin score of candidate is . The candidate with the highest maximin score wins.

Single Transferable Vote (STV): This rule requires up to rounds. In each round, the candidate with the least number of voters ranking them first is eliminated until one of the remaining candidates has a majority.

Bucklin (simplified): The Bucklin score of a candidate is the smallest such that the -approval score of the candidate is strictly larger than . The candidate with the smallest Bucklin score wins.

Note that in some cases, there can be multiple winning candidates (e.g. multiple candidates with the highest Borda score). We therefore may also need a tie-breaking mechanism. All above voting rules can be extended to choose a winner for profiles with weights. In this paper, we study the manipulation problem (with weighted votes), defined as follows.

###### Definition 1

In a manipulation problem, we are given an instance , where is a voting rule, is the non-manipulators’ profile, represents the weights of , is the alternative preferred by the manipulators, is the number of manipulators, and represents the weights of the manipulators. We are asked whether there exists a profile of indivisible votes for the manipulators such that .

When all weights equal to , the problem is called manipulation with unweighted votes. In this paper, we assume that the manipulators controls the tie-breaking mechanism, that is, all ties are broken in favor of .

A large number of normative properties that voting rules might possess have been put forwards including the following.

Unanimity: If a candidate is ranked in the top place by all voters, then this candidate wins.

Monotonicity: If we move the winner up a voter’s preference order, while keeping preferences unchanged, then the winner should not change.

Consistency: If two sets of votes select the same winner then the union of these two sets should also select the same winner.

Majority criterion: If the majority of voters rank a same candidate at the top, then this candidate wins.

Condorcet consistency: If a Condorcet winner exists (a candidate who beats all others in pairwise elections) then this candidate wins.

Condorcet loser criterion: If a Condorcet loser exists (a candidate who is beaten by all others in pairwise elections) then this candidate does not win.

Such properties can be used to compare voting rules. For example, whilst STV satisfies the majority criterion, Borda does not. On the other hand, Borda is monotonic but STV is not.

## 3 Voting Rule Combinator

We consider a simple combinator, written , for combining together two or more voting rules. This combinator collects together the set of winners from the different rules. If all rules agree, this is the overall winner. Otherwise we recursively call the combination of voting rules on this restricted set of winning candidates. If the recursion does not eliminate any candidates, we call some tie-breaking mechanism on the remaining candidates. For example, collects together the plurality and veto winners of an election. If they are the same candidate, then this is the winner. Otherwise, there is a runoff in which we call on the plurality and veto winners. As both plurality and veto on two candidates compute the majority winner, the overall winner of is the winner of a majority election between the plurality and veto winners.

This combinator has some simple algebraic properties. For example, it is idempotent and commutative. That is, and . It has other more complex algebraic properties. For example, . In addition, many normative properties are inherited from the base voting rules. Interestingly, it is sometimes enough for just one of the base voting rules to have a normative property for the composition to have the same property.

###### Proposition 1

For unanimity, the majority criterion, Condorcet consistency, and the Condorcet loser criterion, if one of to and the tie-breaking mechanism satisfy the property, then also satisfy the same property.

On the other hand, there are some properties which can be lost by the introduction of a run-off.

###### Proposition 2 (Monotonicity)

and are both monotonic but is not.

Proof:  Suppose we have 6 votes for , 4 votes for , and 3 votes for and 3 votes for . Tie-breaking for both Borda and plurality is . Now is the Borda winner and is the plurality winner. By tie-breaking, wins the run-off. However, if we modify one vote for to , then becomes the plurality winner and wins the run-off. Hence, is not monotonic.

We give a stronger result for consistency. Scoring rules are consistent, but the combination of any two different scoring rules is not. By “different rules” we mean that there exists a profile for which these two rules select different winners. If two scoring rules are different, then their scoring vectors must be different. We note that the reverse is not true.

###### Proposition 3 (Consistency)

Let and be any two different scoring rules, then is not consistent.

Proof:  Let and be the score given to candidate by and in profile respectively. Since and are different, there exists over to such that on selects and on selects . Then but . WLOG suppose beats in pairwise elections in and tie breaking elects in favour of when they have the same top score. Let consist of votes to where for , ranks in th place and in th place, and ranks in 1st place and in last place. Then and . Let be such that , and be the following profile of cyclic permutations: , , , , , , , . Let be copies of , and be copies of , and copies of . Now on or selects as winner. But on selects .

It follows immediately that is not consistent.

## 4 Strategic Voting

Combining voting rules together can hinder strategic voting. One appealing escape from the Gibbard-Satterthwaite theorem was proposed by Bartholdi, Tovey and Trick . Perhaps it is computationally so difficult to find a successful manipulation that voters have little option but to report their true preferences? As is common in the literature, we consider two different settings: unweighted votes where the number of candidates is large and we have just one or two manipulators, and weighted votes where the number of candidates is small but we have a coalition of manipulators. Whilst unweighted votes are perhaps more common in practice, the weighted case informs us about the unweighted case when we have probabilistic information about the votes . Since there are many possible combinations of common voting rules, we give a few illustrative results covering some of the more interesting cases. With unweighted votes, we prove that computational resistance to manipulation is typically inherited from the base rules. With weighted votes, our results are stronger. We prove that there are many combinations of voting rules where the base rules are polynomial to manipulate but their combination is NP-hard. Combining voting rules thus offers another mechanism to make manipulation more computationally difficult.

### A First Observation

It seems natural that the combination of voting rules inherits the computational complexity of manipulating the base rules. However, there is not a simple connection between the computational complexity of the bases rules and their combination. In this section, we show two examples of artificial voting rules to illustrate this discrepancy. In the first example, manipulation for the base rules are NP-hard, but manipulation for their combination can be computed in polynomial-time; in the second example, manipulation for the base rules are in P, but manipulation for their combination is NP-hard to compute.

###### Proposition 4

There exist voting rules and for which computing a manipulation is NP-hard but computing a manipulation of is polynomial.

Proof:  We give a reduction from 1 in 3-SAT on positive clauses. Boolean variables to are represented by the candidates to . We also have two additional candidates and . Any vote with in first place represents a clause. The first three candidates besides and are the literals in the clause. Any vote with in first place represents a truth assignment. The positive literals in the truth assignment are those Boolean variables whose candidates appear between and in the vote. With 2 candidates, and both elect the majority winner. With 3 or more candidates, elects candidate if there is a truth assignment in the votes that satisfies exactly one out of the three literals in each clause represented by the votes and otherwise elects . Computing a manipulation of is NP-hard. Similarly, with 3 or more candidates, elects candidate if there is a truth assignment in the votes that satisfies exactly one out of the 3 literals in every clause represented by the votes and otherwise elects . Computing a manipulation of is NP-hard. However, is polynomial to manipulate since and always go through to the runoff where the majority candidate wins.

###### Proposition 5

There exist voting rules and for which computing a manipulation is polynomial but computing a manipulation of is NP-hard.

Proof:  The proof uses a similar reduction from 1 in 3-SAT on positive clauses. With 2 candidates, and both elect the majority winner. With 3 or more candidates, elects candidate if there is a truth assignment in the votes that satisfies at least one out of the three literals in each clause represented by the votes and otherwise elects , whilst elects candidate if there is a truth assignment in the votes that satisfies at most one out of the three literals in each clause represented by the votes and otherwise elects . Computing a manipulation of or is polynomial since we can simply construct either the vote that sets each Boolean variable to true or to false. However, computing a manipulation of as it may require solving a 1 in 3-SAT problem on positive clauses.

If computing a manipulation of the base rules is polynomial, it often remains polynomial to compute a manipulation of the combined rules. However, manipulations may now be more complex to compute. We need to find a manipulation of one base rule that is compatible with the other base rules, and that also wins the runoff. We illustrate this for various combinations of scoring rules.

###### Proposition 6

Computing a manipulation of is polynomial.

Proof:   We present a polynomial-time algorithm that checks whether manipulators can make win in the following two steps: we first check for every candidate , whether the manipulators can make to be the plurality winner for while is the veto winner, and beats in the runoff (or ). Then, we check for every candidate whether the manipulators can make to be the veto winner while is the plurality winner, and beats in the runoff.

For the first step, let be a subset of candidates that beat in under . We denote as the difference in the veto score of , , and in . If the veto scores are equal and in the tie-breaking rule then we set . If then can not win under . Otherwise, we place in last positions in exactly manipulator votes. This placing is necessary for to win under . We place in the first position and in the second position in all votes in . We fill the remaining positions arbitrarily. This manipulation is optimal under an assumption that wins under as is always placed in the first position. For each possible candidate for , we check if such a manipulation is possible and check if is the winner of the run-off round. If we find a manipulation we stop. The special case when is analogous.

For the second step, let be the candidate with maximum score in . We denote to be the difference in the plurality scores of and . We place in the first positions in the manipulator votes. The condition is necessary for to win under . We put in the second position in the remaining votes. We put in the second position after in manipulator votes and put in the first position in the remaining votes. To ensure that wins under we perform the same procedure as above. The only simplification is that we do not need to worry about tie-breaking rule as wins tie-breaking by assumption. We fill the remaining positions arbitrarily. This manipulation is optimal under an assumption that wins under , as is placed in the first position unless has to occupy it. For each possible candidate we check if such a manipulation is possible and check if is the winner of the run-off round. If we find a manipulation we stop. Otherwise, there is no manipulation.

It is also in P to decide if a single agent can manipulate an election for any combination of scoring rules. Interestingly, we can use a perfect matching algorithm to compute this manipulation.

###### Proposition 7

Computing a manipulation of is polynomial for a single manipulator and any pair of scoring rules, and .

Proof:  Suppose there is a manipulating vote such that wins under . Let and have the scoring vectors and . As is common in the literature, we assume tie-breaking is in favour of . Suppose wins under in a successful manipulation. The case that wins under is dual. Suppose another candidate wins under , is placed at position and is placed at position in . We show how to construct this vote if it exists by finding a perfect matching in a bipartite graph. For each candidate besides and , we introduce a vertex in the first partition. For each position in we introduce a vertex in the second partition. For each candidate besides and we connect the corresponding vertex with a vertex in the second partition iff (1) the score of in under less the score of in under is less than or equal to , and (2) the score of in under less the score of in under is less than or equal to , or if two differences are equal then is before in the tie-breaking rule. In other words, we look for a placement of the remaining candidates in such that wins in under , wins in under , is at position and is at position in . There exists a perfect matching in this graph iff there is a manipulating vote that satisfies our assumption. If , the reasoning is similar but we only need to fix the position of . Using this procedure, we check for each candidate and for each pair of positions if there exists a vote such that wins in under , wins in under , is at position and is at position in . If such a vote exists, we also check if beats in the run-off round. If loses to in the run-off for all combinations of and then no manipulation exists.

We begin with combinations involving STV. This was the first commonly used voting rule shown to be NP-hard to manipulate by a single manipulator . Not surprisingly, even when combined with voting rules which are polynomial to manipulate like plurality, veto, or -approval, manipulation remains NP-hard to compute.

###### Proposition 8

Computing a manipulation of is NP-hard for for manipulator.

Proof:  (Sketch) Consider the NP-hardness proof for manipulation of STV . We denote the profile constructed in the proof . The main idea is to modify so that the preferred candidate can win under iff can win the modified election under . For reasons of space, we illustrate this for . Other proofs are similar. Candidate (who is the other possible winner of ) has the top Borda score. Hence, must win by winning the STV election (which is possible iff there is a 3-cover). We still have the problem that beats in the run-off. Hence, we introduce a dummy candidate after in each vote. This makes sure that the score of is greater than or equal to . We also introduce blocks of votes. Let . The Borda score of in is greater than that of any other candidate. In an STV election on , reaches the last round. Therefore, the elimination order remains determined by the votes in . Hence, if there is a 3-cover, the candidate can reach the last round. In the worst case, when is divisible by , the plurality scores of and are the same and wins by tie-breaking.

We turn next to combinations of Borda voting, where it is NP-hard to manipulate with two manipulators [6, 3].

###### Proposition 9

Computing a manipulation of by two manipulators is NP-hard for

Proof:  (Sketch) Consider the NP-hardness proof for manipulation of Borda which uses a reduction from the permutation sums problem . Due to the spaces constraint, we consider only . Other proofs are similar but much longer and more complex. The reduction uses the following construction to inflate scores to a desired target. To increase the score of candidate by more than candidates and by more than candidate we consider the following pair of votes:

 ci≻cn≻c1≻…≻cn−1 cn−1≻cn−2≻…≻c1≻ci≻cn

We change the construction by putting in the last place in the first vote in each pair of votes and first place in the second vote, and leaving all other candidates unchanged when we increase the score of by one. This modification does not change the desired properties of these votes. Note that and cannot be winners under . Hence, must win under and then win the run-off. This is possible iff there exists a solution for permutation sums problem.

We focus on elections with weighted votes and 3 candidates. This is the fewest number of candidates which can give intractability. All scoring rules besides plurality (e.g. Borda, veto, 2-approval) are NP-hard to manipulate in this case . We therefore focus on combinations of the voting rules: plurality, cup, Copeland, maximin and Bucklin. Computing a manipulation of each of these rules is polynomial in this case. We were unable to find a proof in the literature that Bucklin is polynomial to manipulate with weighted votes, so we provide one below.

###### Proposition 10

Computing a manipulation of is polynomial with weighted votes and 3 candidates.

Proof:  It is always optimal to place the preferred candidate in the first position as this only decreases the scores of the other 2 candidates, and . We argue that the winner is chosen in one of the first two rounds. In the first round, if still loses to or then there is no manipulation that makes win. In the second round, we must have at least one candidate with a majority. Suppose we did not. Then the sum of scores of the 3 candidates is at most . But the sum of the approval votes is which is a contradiction. Hence, if does not get a majority in this round, one of the other candidates wins regardless of the manipulating votes.

We recall that in this paper all ties are broken in favor of , which is crucial in the proof of the above proposition. In fact, we can show that if some other tie-breaking mechanisms are used, then Bucklin is hard to manipulate with weighted votes, even for candidates.

We next identify several cases where computing a manipulation for combinations of these voting rules is tractable.

###### Proposition 11

Computing a manipulation of , or of is polynomial with weighted votes and 3 candidates.

Proof:  First we consider the outcome of vs and vs assuming that is ranked at the first position by all manipulators.

Case 1. Suppose is a Condorcet loser. In this case, can only win if wins under both and . However, must lose under because never elects the Condorcet loser.

Case 2. Suppose is a Condorcet winner. Then is a winner of both rules as they are both Condorcet consistent.

Case 3. Suppose there exists a candidate such that and even if is ranked first by all manipulators. We argue that if there is a manipulation, then all manipulators can vote . We consider the case that wins under and wins under . The other cases ( wins under , wins under , etc.) are similar. For to win under , all candidates have to have the score of 0 as, by assumption, loses to . Hence, the maximum score of is . Therefore, for to win the following holds and . The only possible agenda is v , and the winner playing . In all other agendas, loses to in one of the rounds. For to win , and tie-breaking has . The manipulation vote will only help achieve the inequalities in both cases.

###### Proposition 12

Computing a manipulation of is polynomial with weighted votes and 3 candidates.

Proof:  We consider three possible outcomes of pairwise comparison between vs and vs assuming that is ranked at the first position by all manipulators.

Case 1. Suppose is a Condorcet loser after the manipulation. can only win overall if wins under both and . However, must lose under .

Case 2. Suppose is a Condorcet winner. Then must be a winner of as this is Condorcet consistent. Hence, regardless of the rest of the manipulating votes, reaches the run-off round and beats any other candidate.

Case 3. Suppose there exists candidate such that and . Note that must guarantee that does not reach the run-off round as loses to in the pairwise elections. There are two sub-cases: wins under and wins under in , or wins under and wins under . As shown in the proof of the last Proposition, if there is a manipulation, will work in both cases.

We continue to focus on combinations of the voting rules: plurality, cup, Copeland, maximin and Bucklin. We give several results which show that there exists combinations of these voting rules where manipulation is intractable to compute despite the fact that all the base rules being combined are polynomial to manipulate. These results provide support for our argument that combining voting rules is a mechanism to increase the complexity of manipulation.

###### Proposition 13

Computing a manipulation of where , is NP-complete with weighted votes and 3 candidates.

Proof:  (Sketch) We consider the case . Other proofs are similar but longer. We reduce from a partition problem in which we want to decide if integers with sum divide into two equal sums of size . Consider the following profile:

 4K a≻b≻c4K a≻c≻b 1K b≻a≻c9K b≻c≻a

For each integer , we have a member of the manipulating coalition with weight . The tie-breaking rule is . The has play , and the winner meets . Note that cannot reach the run-off as they beat in pairwise elections whatever the manipulators do. Note that cannot win the rule. Hence must be the winner. The run-off is , the winner against , the winner (which is the same as the final round of the ). For this to occur, the manipulators have to partition their votes so that exactly manipulators put above and put in the first position (and above ).111Here we abuse the notation by saying “ manipulators”, which we meant “manipulators whose weights sum up to ”. Therefore there exists a manipulation iff there exists a partition. .

###### Proposition 14

Computing a manipulation of where , is NP-complete with weighted votes and 3 candidates.

Proof:  (Sketch) We consider the case . Other proofs are similar but longer. We again reduce from a partition problem. Consider the following profile:

 7K b≻c≻aK b≻a≻c 4K a≻c≻b2K a≻b≻c1   c≻a≻b

For each integer , we have a member of the manipulating coalition with weight . Now, must not reach the run-off round and must win by similar arguments to the last proof. Hence must be the winner. For this to occur, the manipulators have to partition their votes so that exactly manipulators put above , manipulators put in the first position (and above ) and put in the last position in all votes. Therefore there exists a manipulation iff there exists a partition.

###### Proposition 15

Computing a manipulation of where , is NP-complete with weighted votes and 3 candidates.

Proof:  (Sketch) We consider the case . Other proofs are similar but longer. We reduce from a partition problem in which we want to decide if integers with sum divide into two equal sums of size . Consider the following profile:

 4K b≻c≻a2K b≻c≻a 2K a≻b≻c2K a≻c≻b

For each integer , we have a member of the manipulating coalition with weight . Now, must not reach the run-off round and must win by similar arguments to the last proof. Hence must be the winner. For to win , manipulators with total weight at least must rank first. Before the manipulators vote, the score of is , of is and of is . We note that must be ranked above in all manipulators votes and above in manipulators votes, otherwise loses to under . As manipulators must vote , we have , and . This increases the score of to and of to . Now must be ranked above in at least manipulators votes to increase its score to . Hence, the only possible option is if manipulators vote and vote and with weight . In this case the score of all candidates are the same and equal to . By the tie-breaking rule, wins. Therefore, there exists a manipulation iff there exists a partition.

We summarize our results about weighted manipulation in the following table.

## 5 Approximation

One way to deal with the intractability of manipulation is to view computing a manipulation as an approximation problem where we try to minimise the number of manipulators. We argue here that combining voting rules together can make such approximation problems more challenging. In particular, we show that a good approximation method for a rule like will perform very poorly when is combined with a simple rule like or . We consider the algorithm for Borda that computes a manipulation that is within 1 of the optimal number of manipulators .

###### Proposition 16

There exists a family of profiles such that the approximation method on requires manipulators where is an optimum number of manipulators for , is the profile in question and

Proof:  We consider . A similar argument holds for . Consider the following profile: for , and 1 vote for . The tie-breaking rule is where the preferred candidate is . The score of the candidates is , . The algorithm outputs the vote . This increase the number of veto points of by 1. Note that , all have only one veto point. Hence, wins by the tie breaking. Note that has 2 veto-points. However, loses to in the run-off round as is ranked before in votes. Hence, the algorithm will continue to produce pairs, and , until takes first positions in votes and wins the run-off round. On the other hand, if we add the votes for then we increase veto points of candidates by one. Hence, wins under by tie-breaking, and then loses to in the run-off.

## 6 Experimental Results

We investigated the effectiveness of approximation methods on combinations of voting rules experimentally. We used a similar setup to . We generated uniform random votes and votes drawn from the 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. For each combination of the number of candidates , , and the number of voters, , and , we generated 200 instances of elections for each model.

We ran four algorithms. The first algorithm, Opt, finds an optimum solution of the manipulation problem for . We use a constraint solver to encode the manipulation problem as a CSP. We could only solve small problems using complete search as the CSP model is loose. The second algorithm is Greedy algorithm from  that we run until the winner it produces is also a winner of . The third algorithm, Plur, is a greedy algorithm for . Again, we run until its output is a winner of . The fourth algorithm, AdaptGreedy, is our modification of Greedy that simultaneously tries to manipulate and . The algorithm runs the Greedyheuristic first and checks if the preferred candidate is a winner under . If loses under both rules we increase the number of manipulators. If wins under one of the rules we check if we can make a candidate the winner of the other rule, where is the set of candidates. If we want to win under then we place in exactly as many first positions as it needs to win under and place in the remaining first and second positions. We run Greedy to place the remaining positions starting with votes where the first two positions are fixed. If we want to win under , we find the maximum number of first positions for such that still can win under and fill the remaining positions using Greedy. In both case, we check that the preferred candidate is winner under . If not, we increase the number of manipulators. Tables 2 show the results of our experiments. First, they show that we need to adapt approximation algorithms for individual rules to obtain a solution that is close to the optimum number of manipulators. As the size of the election grows individual approximation algorithms require significantly more manipulators than the optimum. Second, for the combination of and , our adaptation of the Greedy method works very well and finds a good approximation. Experimental results suggest that it finds a solution with at most one additional manipulator.

## 7 Conclusion

We have put forwards a simple method for combining together voting rules that performs a run-off between the different winners of each voting rule. We have provided theoretical and experimental evidence for the value of this combinator. On the theoretical side, we proved that a combination of voting rules can inherit a desirable property like Condorcet consistency or the majority criterion from just one base voting rule. On the other hand, two important properties can be lost by the introduction of a run-off: monotonicity and consistency. Combining voting rules also tends to increase the computational difficulty of finding a manipulation. For instance, with weighted votes, we proved that computing a manipulation for a simple combination like and is NP-hard, even though and on their own are polynomial to manipulate. On the experimental side, we studied the impact of this combinator on approximation methods that find close to optimal manipulations.

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