Condorcet ranking: Difference between revisions
no edit summary
No edit summary |
|||
| (3 intermediate revisions by the same user not shown) | |||
Line 1:
A '''Condorcet ranking''' is an ordering of candidates such that every candidate at a given rank [[pairwise beat]]<nowiki/>s every lower-ranked candidate. In other words:
* The [[Condorcet method | Condorcet]] winner, if one exists, is ranked first.
Line 19:
1 C</blockquote>A and B pairwise tie (2 vs. 2), but both beat C (2 vs. 1), so a generalized (or perhaps "weak") Condorcet ranking would be A=B>C.
When there are [[Condorcet cycle|Condorcet cycles]], there won't be a Condorcet ranking (though it may still be possible to fill in certain parts of the [[Order of finish|order of finish]] i.e. if 5 candidates pairwise beat all others, with 3 candidates in a cycle and pairwise beaten by the 5, but otherwise pairwise beating all others, and 2 candidates pairwise beaten by all others except pairwise tied with each other, then the 5 candidates are in the first 5 places of the Condorcet ranking, with the 2 candidates both in 9th place, with the 3 candidates in the cycle guaranteed to be ranked either 6th, 7th, or 8th, depending on which [[Condorcet completion method]] is used).
The most direct generalization of the Condorcet ranking to handle Condorcet cycles is the Smith set ranking, which is meant to ensure that anytime the candidates can be divided into two groups such that everyone in the first group beats everyone in the second group, then everyone in the first group is ranked higher.
The fastest way to find the Condorcet (and Smith) ranking is to find the [[Copeland]] ranking. If every candidate at a given rank has only one more pairwise victory than all candidates at the next-lowest rank (i.e. the number of pairwise victories per candidates incrementally decreases as you go down the order), then there is a Condorcet ranking, otherwise there is only a Smith ranking.
== Smith set ranking ==
[[File:Finding smith set ranking.png|thumb|543x543px|An example of finding the generalized Smith set ranking using the [[Pairwise counting|pairwise matrix]]. One way to better visualize it is to imagine the arrow starting from the top left and traveling down one cell at at a time, creating a box around a group of candidates any time there are only pairwise victories to the right of the cells that are enclosed by the arrow. For further visual clarity, imagine a green background on the cells with pairwise victories; the arrow tries to cover the fewest cells necessary for there to be an all-green background to the right. ]]
[[File:Alternative way to find Smith set ranking.png|thumb|633x633px]]
A '''Smith ranking''' ('''Smith set ranking''') is the same as a Condorcet ranking, except instead of being defined in terms of the Condorcet winner and Condorcet loser, it is defined in terms of the [[Smith set]] and [[Smith set#Smith loser set|Smith loser set]]. When there is a multi-member Smith set, the candidates in the Smith set may be ranked in any order so long as they are all ranked above non-Smith set candidates, with the same applying for the Smith loser set candidates with regards to being ranked lower than non-Smith loser set candidates.<ref>{{Cite web|url=http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.200.8245&rep=rep1&type=pdf|title=AN EXTENSION OF THE CONDORCET CRITERION AND KEMENY ORDERS|last=|first=|date=|website=|url-status=live|archive-url=|archive-date=|access-date=|quote=A first objective of this paper is to propose a formalization of this idea, called the Extended
Condorcet Criterion (XCC). In essence, it says that if the set of alternatives can be partitioned in such a
Line 245 ⟶ 250:
== Notes ==
Smith set rankings can be given additional context by showing for each Smith set the weakest pairwise victory anyone in that Smith set has against anyone in a lower-ranked Smith set; one generalized way to show strength of victory is by % of votes earned in the matchup, which will always be over 50% for the pairwise winner. So, for example, if there are 4 candidates, with 3 of them in a cycle and each of them having a 65%, 60%, and 55% victory respectively against the fourth candidate, then 55% could be used to describe the minimum strength of all of the candidates in the set against the lower-ranked candidate.
| |||