Each possible complete ranking of the candidates is given a "distance" score. For each pair of candidates, find the number of ballots that order them the the opposite way as the given ranking. The distance is the sum across all such pairs. The ranking with the least distance wins.
The winning candidate is the top candidate in the winning ranking.
Imagine that Tennessee is having an election on the location of its capital. The population of Tennessee is concentrated around its four major cities, which are spread throughout the state. For this example, suppose that the entire electorate lives in these four cities, and that everyone wants to live as near the capital as possible.
The candidates for the capital are:
- Memphis, the state's largest city, with 42% of the voters, but located far from the other cities
- Nashville, with 26% of the voters, near the center of Tennessee
- Knoxville, with 17% of the voters
- Chattanooga, with 15% of the voters
The preferences of the voters would be divided like this:
|42% of voters
(close to Memphis)
|26% of voters
(close to Nashville)
|15% of voters
(close to Chattanooga)
|17% of voters|
(close to Knoxville)
Consider the ranking Nashville>Chattanooga>Knoxville>Memphis. This ranking contains 6 orderings of pairs of candidates:
- Nashville>Chattanooga, for which 32% of the voters disagree.
- Nashville>Knoxville, for which 32% of the voters disagree.
- Nashville>Memphis, for which 42% of the voters disagree.
- Chattanooga>Knoxville, for which 17% of the voters disagree.
- Chattanooga>Memphis, for which 42% of the voters disagree.
- Knoxville>Memphis, for which 42% of the voters disagree.
The distance score for this ranking is 32+32+42+17+42+42=207.
It can be shown that this ranking is the one with the lowest distance score (this is because this is the Condorcet ranking, and therefore switching any pair of candidates would require overturning the majority of voters in that pairing rather than the minority). Therefore, the winning ranking is Nashville>Chattanooga>Knoxville>Memphis, and so the winning candidate is Nashville.
Example with a Condorcet cycle:
25 A>B>C 40 B>C>A 35 C>A>B
A>B: 60>40, B>C: 65>35, C>A:75>25. There are 6 main rankings to consider here: A>B>C: A>B opposed by 40, A>C by 75, and B>C by 35. Score is 150. So the minimum score so far is 150. A>C>B: A>C by 75, A>B by 40, C>B by 65. Score is 180. Since this is greater than the minimum (150) this is disqualified. B>A>C: B>A by 60, B>C by 35, A>C by 75. Score is 170. Disqualified by 150. B>C>A: B>C by 35, B>A by 60, C>A by 25. Score is 120. This is the new minimum, so A>B>C is now disqualified. C>A>B: C>A by 25, C>B by 65, A>B by 40. Score is 130. Disqualified by 120. C>B>A: C>B by 65, C>A by 25, B>A by 60. Score is 150. Disqualified by 120.
So the final ranking is B>C>A, with B winning.
If, when the distance score of a ranking A>B>C is being calculated, a voter who ranked B but not A is treated as ordering A and B the opposite way as the ranking, then the Kemeny-Young ranking is a Smith set ranking. This is because any candidate in the n-th Smith set will always be ranked higher than any candidate in a lower Smith set by more voters than vice versa by definition (because the n-th Smith set candidate pairwise beats all candidates in lower Smith sets), so if you take any non-Smith set ranking and minimally modify it to become a Smith set ranking, this will always reduce the distance score. In other words, if there is some ranking which puts a candidate in the n-th Smith set after some candidate in a lower Smith set, then modifying it to swap the two will reduce the distance created by that pair of candidates.
See Pairwise sorted methods, which do what is called "Local Kemenization" to produce a ranking, while being cloneproof.
Some text of this article is derived with permission from Electoral Methods: Single Winner.