DeepAI

# Election Score Can Be Harder Than Winner

Election systems based on scores generally determine the winner by computing the score of each candidate and the winner is the candidate with the best score. It would be natural to expect that computing the winner of an election is at least as hard as computing the score of a candidate. We show that this is not always the case. In particular, we show that for Young elections for dichotomous preferences the winner problem is easy, while determining the score of a candidate is hard. This complexity behavior has not been seen before and is unusual. For example, a common slight variant of Young has a hard winner problem for dichotomous preferences, and natural versions of Dodgson and Kemeny for dichotomous preferences have easy score and winner problems.

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05/18/2020

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## 1 Introduction

Election systems based on scores generally find a winner by computing the score of each candidate and a candidate is a winner if they have the best score. Most election systems have easy score problems and so finding a winner in this way is also easy. The three most-commonly studied election systems where this is not the case are Young [29], Dodgson [8], and Kemeny [17]. The score problem for each of these systems is NP-complete and the corresponding winner problems are each -complete [2, 23, 14, 15].

In general it would seem that determining if a candidate is a winner of an election is at least as hard as determining the score of a candidate. We show that this is not always the case. We show that for Young elections for dichotomous preferences the score problem is hard and the winner problem is easy. This is behavior that was not seen before in the computational study of voting. Dichotomous preferences are a very natural way for voters to state their preferences, where they approve of a subset of the candidates and disapprove of the others. In many situations voters may not be able to state strict preferences over all of the candidates, but can state their approval/disapproval for each candidate. Approval voting is the most well-known election system that uses dichotomous preferences as input. Additionally, most voting rules have definitions that apply or can be naturally extended to dichotomous preferences.

The behavior of Young for dichotomous preferences is quite unusual and does not occur for any of the obvious variations of the setting. When we instead consider strongYoung elections, in which the score is based on how far removed the candidate is from becoming a Condorcet winner (rather than a weak Condorcet winner as in Young), the score and winner problems for dichotomous preferences are hard. This is also the case when the electorate is trichotomous (each voter ranks their most-preferred, middle preferred, and least preferred candidates). Additionally, natural variants of Dodgson and Kemeny for dichotomous preferences do not exhibit Young’s anomalous behavior: Their score and winner problems are in P.

The computational easiness of the winner problem for Young for dichotomous preferences follows from the fact that when the electorate is dichotomous, the majority relation is transitive. To further explore this effect, we consider how having single-peaked preferences and how having single-crossing preferences affects the complexity of the winner and score problems for Young, Kemeny, and Dodgson elections. Single-peaked preferences [4] and single-crossing preferences [20] are the two most-commonly studied domain restrictions that guarantee transitivity of the majority relation and they each model the preferences of the voters with respect to a single polarizing issue. In the case of single-peaked preferences, the axis is a total ordering of the candidates where candidates on the leftmost/rightmost ends of the axis represent the extremes of the issue. In the case of single-crossing preferences, the axis is a total ordering of the voters. We follow the model from Walsh [28] where the axis is given as part of the input.

We consider the complexity of the score and winner problems for Young, strongYoung, Kemeny, Dodgson, and weakDodgson elections for single-peaked and for single-crossing preferences. Except for Dodgson and weakDodgson score for single-crossing preferences, which remain open, all score and winner problems are in P. In proving our results, we solve an open problem from Peters [22] by showing that computing the Dodgson score of a candidate is in P for single-peaked preferences.

## 2 Preliminaries

An election consists of a set of candidates , and a set of voters . Each voter has a corresponding vote (preference order) over the set of candidates. The most commonly studied model is for voters to state their preferences as a total order, i.e., strictly ranking all of the candidates from most to least preferred. Another natural way for voters to state their preferences is as a dichotomous order, where each voter approves of a subset of the candidates and disapproves of the remaining candidates. More formally, given a set of candidates , a dichotomous vote partitions into two sets, and (which may be empty), with the vote written as , such that for all and for all , states (where denotes strict preference) and does not state strict preference between candidates in nor between candidates in .

An election system is a mapping from an election (a set of candidates and a set of voters) to a subset of the candidate set referred to as the winner(s). We are interested in the relationship between the complexity of the score and winner problems for election systems so we focus on election systems with computationally difficult score problems: Young [29], Dodgson [8], and Kemeny [17], and some of their related variants.

For our results and to define the election systems mentioned above, it will be useful to refer to Condorcet winners and weak Condorcet winners. A Condorcet winner is a candidate in an election that beats each other candidate by pairwise majority. Similarly, a weak Condorcet winner is a candidate that beats-or-ties each other candidate by pairwise majority.

The Young score of a candidate is the size of a largest subset of voters for which they are a weak Condorcet winner. A candidate is a Young winner if they have maximum Young score. Notice that the definition of Young applies to dichotomous preferences. We also consider the common slight variant of Young called strongYoung, in which the goal is to make a candidate a Condorcet winner (instead of a weak Condorcet winner).

For total order votes, the Dodgson score of a candidate is the fewest number of swaps between adjacent candidates in the voters’ rankings such that they can become a Condorcet winner. A candidate is a Dodgson winner if they have minimum Dodgson score. The definition for weakDodgson is the same except we consider if a candidate can become a weak Condorcet winner. For dichotomous preferences, a natural analogue for Dodgson is to move candidates between the two groups in a dichotomous vote, which keeps the votes dichotomous, e.g., given the vote we can move up to get with one move.

A Kemeny consensus is a total order with minimal sum of Kendall tau distance to the voters, i.e., , where denotes the number of voters that state . The Kemeny score of a candidate is the minimal sum of Kendall tau distance to the voters for a total order that has as most preferred. For dichotomous preferences, we consider two variants of Kemeny introduced by Zwicker [30] called -Kemeny and -Kemeny. In both systems, the votes are dichotomous.222The definition of dichotomous votes used by Zwicker [30] requires that each of the two groups are nonempty. However, this does not make a difference in the results in this paper. In -Kemeny, the consensus is also dichotomous. In -Kemeny, the consensus is a total order. Having a total order consensus allows expressing more information about the electorate, which sometimes may be more appropriate [1]. In -Kemeny and -Kemeny, the score of ranking is and we are looking to maximize the score.

We now define the score and winner decision problems with Young as an example. Note that in Young we are looking to maximize the score (as is the case for -Kemeny and -Kemeny), but in Dodgson and Kemeny we are looking to minimize the score and so the decision problems for these systems must be adjusted accordingly.

Name: YoungScore

Given: Given an election , a candidate , and a number .

Question: Is the Young score of at least ?

Name: YoungWinner

Given: Given an election and a candidate .

Question: Does have highest Young score?

Our computational results involve the classes P, NP, and . The class was first studied by Papadimitriou and Zachos [21], named by Wagner [27], and shown by Hemachandra [13] to be equivalent to , the class of problems solvable by a polynomial-time oracle machine that asks all of its queries to an NP oracle in parallel. Note that .

## 3 Dichotomous Preferences

For total order votes, the winner problems for Young, strongYoung, Dodgson, weakDodgson, and Kemeny are -complete [5, 23, 14, 5, 15]. The upper bounds for these problems are shown as follows: Use the associated NP-complete score problem, and compute the scores of all candidates in parallel in polynomial time.

An election system is weakCondorcet-consistent if on every input that has at least one weak Condorcet winner, the winners of the election system are exactly the weak Condorcet winners. If an election has dichotomous votes then it has at least one weak Condorcet winner [16], which implies the following theorem.

###### Theorem 1

The winner problem for a weakCondorcet-consistent election system with dichotomous preferences is in P. This holds even for election systems that are weakCondorcet-consistent when restricted to dichotomous preferences.

### 3.1 Young Elections

###### Theorem 2

For dichotomous preferences, YoungWinner is in P and YoungScore is NP-complete.

Proof.    Since Young is weakCondorcet-consistent [10], YoungWinner in P follows from Theorem 1.

We will now show that YoungScore remains NP-complete for dichotomous preferences. And so the “natural” way of deciding YoungWinner by using YoungScore as an oracle is not optimal.

We reduce from Independent-Set. Given a graph , let the candidate set be and let the voter set consist of the following voters.

• For each vertex , one voter corresponding to voting .

• One voter voting .

Note that the Young score of is , where is the independence number of , i.e., the size of a maximum independent set of . This score is realized by the voter that ranks first and a set of voters corresponding to a maximum-size independent set of . Note that the proof of the current theorem does not contradict the previous statement that YoungWinner is in P, since is clearly in general not a Young winner. Also note that this construction does not give NP-completeness for YoungScore for total orders. Not surprisingly, it turns out that that problem is also NP-complete (even to approximate [6]). Somewhat surprisingly, a direct proof of NP-completeness is not given in the literature, but it is implicit in the proof of Rothe, Spakowski, and Vogel [23].333It is interesting to note that the NP-completeness of YoungScore does not directly follow from the -completeness of YoungWinner and strongYoungWinner: Under the assumption that NP does not have p-measure 0, there exists a set that is NP-complete under truth-table reductions, but not NP-complete (under many-one reductions) [18]. Note that YoungWinner and strongYoungWinner are in , though is not NP-complete.

It may be surprising that the construction for total orders is harder than the one for dichotomous preferences, because it may seem that a problem on total orders would be harder than the analogous problem for dichotomous preferences. However, for the plurality rule, control by adding voters can be NP-complete for votes with ties [11] while it is in P for total orders [3]. And it is easy to see that dichotomous votes suffice to get NP-completeness.   ❑

###### Theorem 3

strongYoungScore for dichotomous preferences is NP-complete.

We now consider the ranking problem, which is the problem that asks given two candidates and whether the score of is at least the score of . This will be used as an intermediate problem to show the hardness of strongYoungWinner and for Young, to further contrast the complexity of the score, ranking, and winner problems. The construction also shows that the YoungLoser problem is -complete, in stark contrast to the YoungWinner problem.

###### Theorem 4

YoungRanking for dichotomous preferences is -complete.

Proof.    Reduce from Min-Card-Independent-Set-Compare, in which we are given two graphs and with the same number of vertices and we ask if . This problem is -complete [26] (for an explicit proof, see [25]).

Without loss of generality, assume that and both contain at least one edge (so that and ) and that the sets of vertices are disjoint. Our reduction generalizes the YoungScore reduction from the proof of Theorem 2. We will ensure that the Young score of is and that the Young score of is . This proves the theorem, since .

Let the candidate set be and let the voter set consist of the following voters.

Type I

For each , one voter corresponding to voting .

Type II

One voter voting .

Type III

For each , one voter corresponding to voting .

Type IV

One voter voting .

Note that to realize the Young score of , we should always include all the Type III and Type IV voters. The rest of the argument is as in the proof of Theorem 2. Note that ties-or-beats each candidate in , since we are including Type I votes.   ❑

Add a vote and a vote to get the following.

###### Theorem 5

strongYoungRanking for dichotomous preferences is -complete.

As stated in Theorem 2, YoungWinner is in P for dichotomous votes. In contrast, the winner problem for strongYoung for dichotomous votes is -complete.

###### Theorem 6

strongYoungWinner for dichotomous preferences is -complete.

Proof.    The main insight here is that we can always make sure that a candidate ’s strongYoung score is 0, by adding a candidate and making sure that and are tied in every vote. We adapt the construction used to show that strongYoungRanking for dichotomous preferences is -complete from Theorem 5. For each , add a candidate and make sure that and are tied in every vote.   ❑

It is interesting to see that our -completeness proofs for dichotomous preferences are also significantly easier than those for total orders [23].

Note that the approach from Theorem 6 does not work for YoungWinner. Rothe, Spakowski, and Vogel [23] reduce the strongYoungRanking to strongYoungWinner problem by replacing each candidate other than and by new candidates , and by replacing the occurrence of in the th voter by (modulo ). This does not change the scores of and , but ensures that the strongYoung score of every other candidate is at most 1, and so we can ensure that these candidates are never winners in the image of the reduction. Note that this construction does not work for dichotomous preferences (which is consistent with the Theorem 2 result that YoungWinner for dichotomous preferences is in P). The construction also does not work for trichotomous preferences, or indeed for any -chotomous preferences.

However, we can adapt the construction used to show that YoungRanking for dichotomous preferences is -complete from Theorem 4. For details, see the appendix.

###### Theorem 7

YoungWinner for trichotomous preferences is -complete.

### 3.2 Dodgson Elections

We now show that the complexity behavior of Young for dichotomous preferences does not occur for Dodgson and Kemeny for dichotomous preferences.

Recall that for Dodgson for dichotomous preferences a candidate can only be swapped up from disapproved to approved or swapped down from approved to disapproved. So unlike in the case for total orders, to determine the score of a candidate we cannot just consider swaps that move that candidate up. For example, given the vote , for to beat pairwise, we first need to move up to get and then move down to get . (We could of course also do these moves in the reverse order.) With this in mind, we can show that DodgsonScore and weakDodgsonScore are each in P (and so are the corresponding winner problems).

###### Theorem 8

DodgsonScore and weakDodgsonScore, for dichotomous preferences, are each in P.

Proof.    First notice that moving up from the set of disapproved candidates in a vote to the set of approved candidates decreases by 1 for every candidate . And it is not possible to decrease by more than 1 with one move. So, the weakDodgson score of is the max over all candidates of : We can make a weak Condorcet winner by moving up from disapproved to approved in that many votes and it is easy to see that there are enough votes to do this. So, weakDodgsonScore is in P.

For DodgsonScore, if starts out as a Condorcet winner, the score is 0. Otherwise, we need to move up in one more vote compared to what was needed to make a weak Condorcet winner. If no such vote exists then for some candidate , in each original vote. In this case, move up in every vote, and then for every candidate for which and are tied, move that candidate down. So, DodgsonScore is also in P.   ❑

### 3.3 Kemeny Elections

Zwicker [30] shows that the winner problems for -Kemeny and -Kemeny are in P. An easy way to see this for -Kemeny is because this is the same as the mean rule [30]. Computing the score of a candidate is a little harder, since we need to rank (tied for) first in a dichotomous ranking, and so we are not ranking according to approval score (as in the mean rule). However, a similar argument to what is used in the proof of Lemma 3 from Zwicker [30], which computes a -chotomous consensus, can be used to show that the score problem for -Kemeny is also in P.

For Kemeny versions that require a total order consensus (such as -Kemeny) we will show that the score problem polynomial-time Turing-reduces to the winner problem, which implies that the score problem is in P if the winner problem is in P. And it follows that -KemenyScore is in P.

###### Theorem 9

For each preference domain that is closed when candidates are removed, KemenyScore for -preferences polynomial-time Turing-reduces to KemenyWinner for -preferences with a total order consensus.

Proof.    The reduction in the theorem above works in the following way. To compute the Kemeny score of candidate , we need to compute the score of a total order that ranks first. So, put first in the total order. The contribution of to the score is independent of how the remaining candidates are ordered (for example, contributes for -Kemeny). For an optimal order, we need to order such that the order restricted to those candidates is optimal, i.e., we need to compute a Kemeny consensus of the electorate restricted to .

So, delete and repeatedly query whether a candidate is a winner. If so, put next in the order and delete . This builds an optimal total order with first. It is clear this argument holds for preference domains that are closed under removal of candidates and this includes total order, dichotomous, single-peaked, and single-crossing preferences.   ❑

## 4 Single-Peaked Preferences

Single-peaked preferences model the preferences of the electorate with respect to a one-dimensional axis , a total ordering of the candidates, where each voter has a single most preferred candidate (their peak) and candidates farther to the leftmost (rightmost) ends of the axis are strictly less preferred. More formally, for every triple of candidates or , for every voter if states then states .

The pairwise majority relation for single-peaked preferences is transitive. Brandt et al. [5] show that it follows from their Theorem 3.2 (the analogue of Theorem 1) that the winner problems for Kemeny, Young, and weakDodgson are in P. They also show that the winner problems for strongYoung and Dodgson are in P.

So, what happens to the single-peaked score problems of the systems mentioned above?

#### Dodgson Elections

Peters [22] states that “… while Brandt et al. (2015) give an efficient algorithm for finding a Dodgson winner in the case of single-peaked preferences, the problem of efficiently calculating scores appears to be open and non-trivial.” The reason for this nontriviality is that after swapping, the electorate is not required to be single-peaked (see [5, Footnote 5]). This makes single-peaked Dodgson very different from single-peaked Kemeny and Young, where we will never get non-single-peaked electorates in the computation. We will show below that for Young and Kemeny, the single-peaked score problem is in P.

More surprisingly, we also show the following theorem.

###### Theorem 10

DodgsonScore and weakDodgsonScore for single-peaked preferences are in P.

Proof.    We start with an example that shows the core argument of our algorithm. Suppose the societal axis is and suppose our election consists of the following types of votes.

10 votes of the form .

50 votes of the form .

10 votes of the form .

20 votes of the form .

11 votes of the form .

In order for to become a Condorcet winner, every candidate other than can be preferred to by at most 50 voters. does not need any votes over , and we leave the 10 votes of the form as is. needs 10 votes over . In 10 of the votes of the form , we swap to the top of the order (using 30 swaps total) and we leave the remaining 40 votes of this form as is. Note that in the resulting set of votes:

10 votes of the form .

40 votes of the form .

10 votes of the form .

, , and are preferred to by 50 voters. This implies that we need to swap to the top of the preference order in all votes of the form and all votes of the form . Note that no swap is wasted, and so has a Dodgson score of 70.

We now show that this approach can be generalized to all single-peaked electorates. Let the societal axis be . Our goal is to compute the (Dodgson or weakDodgson) score of . Let (for “half”) be the maximum number of voters that can prefer to while making a winner. To be precise, in the case of Dodgson winner, is , and in the case of weakDodgson winner, is , where is the number of voters. In the example above, is 50. We will show that we can make a winner without wasting swaps, i.e., in the election that makes a winner, is preferred to by exactly voters for every candidate such that , and by voters for every candidate such that . Since we are not wasting swaps, this witnesses the Dodgson score of (with value ).

Note that in each single-peaked vote, is preferred to all candidates or to all candidates. So, when we swap up in a vote, this will help against candidates or against candidates, but not against both. This implies that we can treat the candidates and candidates separately. We will show that we can ensure that each candidate is preferred to by at most voters without wasting swaps. The same holds for the candidates, which proves the theorem.

As in the example, we classify the voters in terms of the set of

candidates that are preferred to . For every , let be the number of voters with a vote of the form and let be the number of voters that prefer to all candidates. So, , the total number of voters is , and is defined as above.

If for all , , then does not need to gain any points over candidates. Otherwise, let be the smallest index such that . We now swap to top of the preference order of voters with a vote of the form .444Note that . This is immediate if . If , , and thus . Also, for all , we swap to the top of the preference order of all the voters with a vote of the form .

We now show that after swapping, each candidate is preferred to by at most voters without wasting swaps. That is, is preferred to by voters for all and is preferred to by voters for all .

For , since we never swap over , it is immediate that is preferred to by the same voters as in the original election. For , is preferred to by all unchanged voters that prefer to . And there are exactly such voters.   ❑

The proof of the above theorem shows that for single-peaked electorates, the Dodgson score of is equal to . This also gives a simple algorithm for DodgsonWinner. Though this problem was known to be in P [5], that algorithm was more complicated, since it did not look at the form of the entire single-peaked electorate in the way we do in our proof.

#### Young Elections

It is easy to see that we can adapt the construction for Dodgson from the previous section to Young. For every voter that we swap to the top of the preference order of in the Dodgson construction, we now delete that voter. It is easy to see that this gives the minimum number of voters to delete in order for to become a Condorcet (weak Condorcet) winner, and so the number of remaining voters is exactly the strongYoung (Young) score of . This result also follows from the result that constructive control by deleting voters for Condorcet and weak Condorcet elections are each in P [5].

#### Kemeny Elections

Brandt et al. [5] show that KemenyWinner for single-peaked preferences is in P. Since we are computing a total order consensus, it follows from Theorem 9 that KemenyScore for single-peaked preferences Turing-reduces to KemenyWinner for single-peaked preferences, and is thus also in P.

## 5 Single-Crossing Preferences

Another important domain restriction that ensures the majority relation is transitive is the single-crossing restriction [20], where the voters can be ordered along a one-dimensional axis such that for each pair of candidates all of the voters that state precede the voters that state , i.e., there is a single crossing point for each pair.

As in the case for dichotomous preferences and for single-peaked preferences, we immediately obtain that YoungWinner and weakDodgsonWinner are each in P. It follows from Magiera and Faliszewski [19] that strongYoungScore and strongYoungWinner are each in P. As mentioned by Cornaz et al. [7], KemenyWinner for single-crossing elections is in P. This immediately implies using Theorem 9 that KemenyScore is in P.

An important property for the complexity is the fact that single-crossing elections have a variant of the median voter theorem, in that the median voter(s) represent the majority relation [24, 12]. The following corollary extracts the properties that we need.

###### Corollary 11

Let be a single-crossing order of voters. A candidate is a weak Condorcet winner if and only if

• is odd and

is the most preferred candidate of the median voter.

• is even and for every , is preferred to by at most one of the two median voters (and by exactly one of the median voters if and only if and are tied).

###### Theorem 12

YoungScore for single-crossing preferences is in P.

Proof.    For every voter that has at the top of its preference order, keep a maximum odd number of voters such that is the median voter. If the number of remaining voters is greater than the current best score, this becomes the current best score.

For every pair of voters and , if there is a candidate that is preferred to by and , go to the next loop iteration. Otherwise, keep a maximum even number of voters such that and are the two median voters. If the number of remaining voters is greater than the current best score, this becomes the current best score.   ❑

###### Theorem 13

DodgsonWinner for single-crossing preferences is in P.

Proof.    This is trivial if there are an odd number of voters. So, assume that the number of voters is even. We will show that every Dodgson winner is a weak Condorcet winner, and that the Dodgson scores of weak Condorcet winners are easy to compute. This immediately implies the theorem, since the Dodgson winners are the weak Condorcet winners with lowest Dodgson score.

Suppose is a weak Condorcet winner. It follows from Corollary 11 that for every candidate , is preferred to by at least one of the median voters and is preferred to by one of the median voters if and only if is tied with pairwise. To make a Condorcet winner with a minimal number of swaps, it suffices to swap to the top of the two median voters. This gives a Dodgson score of .

If is not a weak Condorcet winner, then there is a weak Condorcet winner such that beats pairwise. It is easy to see that the Dodgson score of is greater than the Dodgson score of , since for every candidate , if needs a vote over , then so does . In addition, needs two votes over . It follows that the Dodgson score of is greater than which is the Dodgson score of .   ❑

It is interesting to see that the algorithm for DodgsonWinner for single-crossing preferences from the previous theorem is similar to the algorithm for single-peaked preferences from Brandt et al. [5], in that there it also was shown that only weak Condorcet winners can be winners and how to compute the Dodgson score of a weak Condorcet winner. In Theorem 10, we finally solved the open problem of computing the Dodgson and weakDodgson scores for single-peaked preferences. We have not managed to solve the complexity of these problems for single-crossing electorates. Recall that it was crucial that in the single-peaked case we could always realize the score without wasting swaps. That is not the case in the single-crossing case, as shown by the following simple example. This gives some indication that the single-crossing case may be harder to handle.

###### Example 14

Consider the following four voters, single-crossing w.r.t. the ordering .

voting .

voting .

voting .

voting .

To become a Condorcet winner needs one vote over , one vote over , and three votes over . However, the three votes over can only be obtained by wasting an extra swap over either or . So the Dodgson score of is six.

## 6 Future Work

Our major open question is the complexity of Dodgson score for single-crossing elections. Using the phrasing of Peters [22]: while we give an efficient algorithm for finding a Dodgson winner in the case of single-crossing preferences, the problem of efficiently calculating scores appears to be open and nontrivial.

We also point out that there are other options for “dichotomous Kemeny” depending on the amount a tie contributes to the distance. Kemeny [17] adds 0.5 and Fagin et al. [9] considers all penalty values between 0 and 1. It should be noted that -Kemeny does not correspond to any of these penalty values, since this system depends solely on the induced majority graph, whereas the others do not. Computing a dichotomous consensus for is a very interesting challenge. (Computing a total order consensus is equivalent to -Kemeny.)

Acknowledgments: This work was supported in part by NSF-DUE-1819546, a Renewed Research Stay grant from the Alexander von Humboldt Foundation, and NSF Graduate Research Fellowship DGE-1102937. We thank Lane Hemaspaandra for helpful comments.

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## Appendix

### Proof of Theorem 7: YoungWinner for trichotomous preferences is Θp2-complete.

Proof.    The main insight is that we can make sure that candidate ’s Young score is relatively low, by adding candidates and , tied with in each original vote, and adding, for some large integer , the following voters.

• voters voting .

• voters voting .

• voters voting .

• voters voting .

• voters voting .

• voters voting .

Note that the Young scores of , , and are at most , where is the number of voters.

We adapt the construction used to show that YoungRanking is -hard from Theorem 4. For each , we add and , we replace each occurrence of in the original votes by , and we add voters as specified above, taking . For readability, we write for . This gives the following set of voters.

Type I

For each vertex ,

• One voter corresponding to voting
.

Type II
• One voter voting .

Type III

For each vertex ,

• One voter corresponding to voting
.

Type IV
• One voter voting .

Type V

For each ,

• voters voting .

• voters voting .

• voters voting .

• voters voting .

• voters voting .

• voters voting .

It is immediate that for all , . We will show that and . Since , it follows that if and only if is a Young winner, which completes the proof.

We will show that . follows by symmetry.

Let be a set of voters that realizes the Young score of with a maximal number of Type V voters. Note that all the voters that rank tied for first are in . So, all Type II, III, and IV voters, and half of the Type V voters are in .

Next, we will show that contains all Type V voters. Suppose it does not. Then there is an such that there is a voter voting or or that is not in . Since contains a maximal number of Type V voters, adding to will cause to not be a weak Condorcet winner. It follows that contains a Type I voter that prefers , , and to . But note that is still a weak Condorcet winner in if we replace by (since does not do worse against any candidate in compared to ). This contradicts the assumption that contains a maximal number of Type V voters.

So, contains all Type II, III, IV, and V voters. Then contains Type I voters corresponding to a maximum independent set of (recall that ).   ❑