Editing Condorcet method (section)

==Single-method systems==
{{More citations needed section|date=March 2021}}
Some Condorcet methods use a single procedure that inherently meets the Condorcet criteria and, without any extra procedure, also resolves circular ambiguities when they arise. In other words, these methods do not involve separate procedures for different situations. Typically these methods base their calculations on pairwise counts. These methods include:

*[[Copeland's method]]: This simple method involves electing the candidate who wins the most pairwise matchings. However, it often produces a tie.
*[[Kemeny–Young method]]: This method ranks all the choices from most popular and second-most popular down to least popular.
* [[Minimax Condorcet|Minimax]]: Also called ''Simpson'', ''Simpson–Kramer'', and ''Simple Condorcet'', this method chooses the candidate whose worst pairwise defeat is better than that of all other candidates. A refinement of this method involves restricting it to choosing a winner from among the Smith set; this has been called ''Smith/Minimax''.
* [[Nanson's method]] and [[Baldwin's method]] combine Borda Count with an instant runoff procedure.
* [[Dodgson's method]] extends the Condorcet method by swapping candidates until a Condorcet winner is found. The winner is the candidate which requires the minimum number of swaps.
* [[Ranked pairs]] breaks each cycle in the pairwise preference graph by dropping the weakest majority in the cycle, thereby yielding a complete ranking of the candidates. This method is also known as ''Tideman'', after its inventor [[Nicolaus Tideman]].
*[[Schulze method]] iteratively drops the weakest majority in the pairwise preference graph until the winner becomes well defined. This method is also known as ''Schwartz sequential dropping'' (SSD), ''cloneproof Schwartz sequential dropping'' (CSSD), ''beatpath method'', ''beatpath winner'', ''path voting'', and ''path winner''.
*Smith Score is a rated voting method which elects the Score voting winner from the Smith set.<ref>{{Cite web |title=Smith Score - electowiki |url=https://electowiki.org/wiki/Smith//Score |access-date=2024-12-14 |website=electowiki.org |language=en}}</ref>

Ranked Pairs and Schulze are procedurally in some sense opposite approaches (although they very frequently give the same results):
* Ranked Pairs (and its variants) starts with the strongest defeats and uses as much information as it can without creating ambiguity.
* Schulze repeatedly removes the weakest defeat until the ambiguity is removed.

Minimax could be considered as more "blunt" than either of these approaches, as instead of removing defeats it can be seen as immediately removing candidates by looking at the strongest defeats (although their victories are still considered for subsequent candidate eliminations). One way to think of it in terms of removing defeats is that Minimax removes each candidate's weakest defeats until some group of candidates with only pairwise ties between them have no defeats left, at which point the group ties to win.{{cn|date=March 2024}}

===Kemeny–Young method===
{{Unreferenced section|date=March 2021}}
{{Main|Kemeny–Young method}}
The Kemeny–Young method considers every possible sequence of choices in terms of which choice might be most popular, which choice might be second-most popular, and so on down to which choice might be least popular. Each such sequence is associated with a Kemeny score that is equal to the sum of the [[#Pairwise counting and matrices|pairwise counts]] that apply to the specified sequence. The sequence with the highest score is identified as the overall ranking, from most popular to least popular.

When the pairwise counts are arranged in a matrix in which the choices appear in sequence from most popular (top and left) to least popular (bottom and right), the winning Kemeny score equals the sum of the counts in the upper-right, triangular half of the matrix (shown here in bold on a green background).

{| class="wikitable" style="text-align:center"
!
! ...over '''Nashville'''
! ...over '''Chattanooga'''
! ...over '''Knoxville'''
! ...over '''Memphis'''
|-
! Prefer '''Nashville'''...
| —
| style="background:#cfc;"| '''68'''
| style="background:#cfc;"| '''68'''
| style="background:#cfc;"| '''58'''
|-
! Prefer '''Chattanooga'''...
| 32
| —
| style="background:#cfc;"| '''83'''
| style="background:#cfc;"| '''58'''
|-
! Prefer '''Knoxville'''...
| 32
| 17
| —
| style="background:#cfc;"| '''58'''
|-
! Prefer '''Memphis'''...
| 42
| 42
| 42
| —
|}

In this example, the Kemeny Score of the sequence Nashville > Chattanooga > Knoxville > Memphis would be 393.

Calculating every Kemeny score requires considerable computation time in cases that involve more than a few choices. However, fast calculation methods based on [[integer programming]] allow a computation time in seconds for some cases with as many as 40 choices.

===Ranked pairs===
{{Main|Ranked pairs}}
{{Unreferenced section|date=March 2021}}

The order of finish is constructed a piece at a time by considering the (pairwise) majorities one at a time, from largest majority to smallest majority. For each majority, their higher-ranked candidate is placed ahead of their lower-ranked candidate in the (partially constructed) order of finish, except when their lower-ranked candidate has already been placed ahead of their higher-ranked candidate.

For example, suppose the voters' orders of preference are such that 75% rank B over C, 65% rank A over B, and 60% rank C over A. (The three majorities are a [[rock paper scissors]] cycle.) Ranked pairs begins with the largest majority, who rank B over C, and places B ahead of C in the order of finish. Then it considers the second largest majority, who rank A over B, and places A ahead of B in the order of finish. At this point, it has been established that A finishes ahead of B and B finishes ahead of C, which implies A also finishes ahead of C. So when ranked pairs considers the third largest majority, who rank C over A, their lower-ranked candidate A has already been placed ahead of their higher-ranked candidate C, so C is not placed ahead of A. The order of finish is "A, B, C" and A is the winner.

An equivalent definition is to find the order of finish that minimizes the size of the largest reversed majority. (In the 'lexicographical order' sense. If the largest majority reversed in two orders of finish is the same, the two orders of finish are compared by their second largest reversed majorities, etc. See [[Lexicographical order|the discussion of MinMax, MinLexMax and Ranked Pairs in the 'Motivation and uses' section of the Lexicographical Order article]]). (In the example, the order of finish "A, B, C" reverses the 60% who rank C over A. Any other order of finish would reverse a larger majority.) This definition is useful for simplifying some of the proofs of Ranked Pairs' properties, but the "constructive" definition executes much faster (small polynomial time).

===Schulze method===
{{Main|Schulze method}}

The [[Schulze method]] resolves votes as follows:

:At each stage, we proceed as follows:

:# For each pair of undropped candidates X and Y: If there is a directed path of undropped links from candidate X to candidate Y, then we write "X → Y"; otherwise we write "not X → Y".
:# For each pair of undropped candidates V and W: If "V → W" and "not W → V", then candidate W is dropped and all links, that start or end in candidate W, are dropped.
:# The weakest undropped link is dropped. If several undropped links tie as weakest, all of them are dropped.

:The procedure ends when all links have been dropped. The winners are the undropped candidates.

In other words, this procedure repeatedly throws away the weakest pairwise defeat within the top set, until finally the number of votes left over produce an unambiguous decision.

===Defeat strength===
{{More citations needed section|date=March 2021}}
Some pairwise methods—including minimax, Ranked Pairs, and the Schulze method—resolve circular ambiguities based on the relative strength of the defeats. There are different ways to measure the strength of each defeat, and these include considering "winning votes" and "margins":

*Winning votes: The number of votes on the winning side of a defeat.
*Margins: The number of votes on the winning side of the defeat, minus the number of votes on the losing side of the defeat.<ref>{{cite web|url=https://principles.liquidfeedback.org/The_Principles_of_LiquidFeedback_1st_edition_online_version.pdf |title=The Principles of Liquid Feedback|edition=1 |first1=Jan |last1=Behrens |first2=Axel |last2=Kistner |first3=Andreas |last3=Nitsche |first4=Bjorn |last4=Swierczek |year= 2014}}</ref>

If voters do not rank their preferences for all of the candidates, these two approaches can yield different results. Consider, for example, the following election:

{| class="wikitable"
|-
!45 voters
!11 voters
!15 voters
!29 voters
|-
|1. A
|1. B
|1. B
|1. C
|-
|
|
|2. C
|2. B
|}

The pairwise defeats are as follows:

*B beats A, 55 to 45 (55 winning votes, a margin of 10 votes)
*A beats C, 45 to 44 (45 winning votes, a margin of 1 vote)
*C beats B, 29 to 26 (29 winning votes, a margin of 3 votes)

Using the winning votes definition of defeat strength, the defeat of B by C is the weakest, and the defeat of A by B is the strongest. Using the margins definition of defeat strength, the defeat of C by A is the weakest, and the defeat of A by B is the strongest.

Using winning votes as the definition of defeat strength, candidate B would win under minimax, Ranked Pairs and the Schulze method, but, using margins as the definition of defeat strength, candidate C would win in the same methods.

If all voters give complete rankings of the candidates, then winning votes and margins will always produce the same result. The difference between them can only come into play when some voters declare equal preferences amongst candidates, as occurs implicitly if they do not rank all candidates, as in the example above.

The choice between margins and winning votes is the subject of scholarly debate. Because all Condorcet methods always choose the Condorcet winner when one exists, the difference between methods only appears when cyclic ambiguity resolution is required. The argument for using winning votes follows from this: Because cycle resolution involves disenfranchising a selection of votes, then the selection should disenfranchise the fewest possible number of votes. When margins are used, the difference between the number of two candidates' votes may be small, but the number of votes may be very large—or not. Only methods employing winning votes satisfy [[plurality criterion|Woodall's plurality criterion]].

An argument in favour of using margins is the fact that the result of a pairwise comparison is decided by the presence of more votes for one side than the other and thus that it follows naturally to assess the strength of a comparison by this "surplus" for the winning side. Otherwise, changing only a few votes from the winner to the loser could cause a sudden large change from a large score for one side to a large score for the other. In other words, one could consider losing votes being in fact disenfranchised when it comes to ambiguity resolution with winning votes. Also, using winning votes, a vote containing ties (possibly implicitly in the case of an incompletely ranked ballot) does not have the same effect as a number of equally weighted votes with total weight equaling one vote, such that the ties are broken in every possible way (a violation of Woodall's symmetric-completion criterion), as opposed to margins.<ref>{{Cite journal |first=D R |last=Woodall |title=Properties of Preferential Election Rules |journal=Voting Matters |issue =3 |pages= 8–15 |url=http://www.mcdougall.org.uk/VM/ISSUE3/P5.HTM |access-date=2024-12-14}}</ref>

Under winning votes, if two more of the "B" voters decided to vote "BC", the A->C arm of the cycle would be overturned and Condorcet would pick C instead of B. This is an example of "Unburying" or "Later does harm". The margin method would pick C anyway.

Under the margin method, if three more "BC" voters decided to "bury" C by just voting "B", the A->C arm of the cycle would be strengthened and the resolution strategies would end up breaking the C->B arm and giving the win to B. This is an example of "Burying". The winning votes method would pick B anyway.