[EM] non-transitive ranked pairs

[Ross Hyman]

Ranked Pairs and all of its variants that I am aware of abhor
non-transitivity.  Here is a variant of
Ranked Pairs that embraces non-transitivity. 
Despite being non-transitive, it elects a unique winner, the Condorcet
winner if there is one.  In the cases I
have looked at, the winner is also the Schulze winner.  Is it always?

 

Candidates are classed in two categories: Winners and Losers.  Initially, all candidates are Winners.  Every candidate has an associated List of
candidates that have defeated it.  Every
candidate initially has a List composed of itself and no other candidates.  The method is so affirming of
non-transitivity that it even treats each candidate as a non-transitivity loop
unto itself.  Winners are those
candidates who have no Winners in their List aside from themselves. 

 

Rank the pairs in a strict order, in the same order one
would use for your favorite strict order transitive variant of Ranked Pairs.  Affirm each pair in order, from highest
ranked to lowest.  When A > B is
affirmed, the List for candidate A is added to every List that includes candidate
B (not just candidate B’s list).  All
Winners that now have other Winners in their List are reclassified as
Losers.  The count can be ended when only
one Winner remains since affirming the remaining pairs cannot make the Winner a
Loser.  Provided that every pair is
ranked into a strict ranking, and each pair expresses a definite ranking
between the two candidates in the pair, there is guaranteed to be one Winner.

 

Example election from:
http://www.cs.wustl.edu/~legrand/rbvote/desc.html 

Brad > Erin
623, 298

Erin > Dave 610, 311

Dave > Brad 609, 312

Abby > Erin
511, 410

Abby > Dave 485, 436

Brad > Abby 463, 458

Abby > Cora 461, 460

Brad > Cora 461, 460

Dave > Cora 461, 460

Erin > Cora 461, 460

 

Each Candidate begins as a Winner with only
itself on its List.

Abby(W): Abby(W)

Brad(W):Brad(W)

Cora(W): Cora(W)

Dave(W): Dave(W)

Erin(W): Erin(W)

 

The first affirmed pair is Brad> Erin.  Brad’s List is
added to Erin’s List, the only one that includes Erin.

Abby(W): Abby(W)

Brad(W):Brad(W)

Cora(W): Cora(W)

Dave(W): Dave(W)

Erin(L): Erin(L), Brad(W)

Erin is now a Looser.

 

The next pair to be affirmed is Erin > Dave.  Erin’s List is added to Dave’s List, the only one that
includes Dave.

Abby(W): Abby(W)

Brad(W):Brad(W)

Cora(W): Cora(W)

Dave(L): Dave(L), Erin(L), Brad(W)

Erin(L): Erin(L), Brad(W)

Dave is now a Looser.

 

The next pair to be affirmed is Dave >
Brad.  Dave’s List is added to Brad’s and
Erin’s Lists, since both include Brad.

Abby(W): Abby(W)

Brad(W):Brad(W), Dave(L), Erin(L)

Cora(W): Cora(W)

Dave(L): Dave(L), Erin(L), Brad(W)

Erin(L): Erin(L), Brad(W), Dave(L)

Brad is still a Winner.

 

The next pair to be affirmed is Abby > Erin.  Abby’s List
is added to Brad’s, Dave’s, and Erin’s Lists, since they all include Erin.  

Abby(W): Abby(W)

Brad(L):Brad(L), Dave(L), Erin(L), Abby(W)

Cora(W): Cora(W)

Dave(L): Dave(L), Erin(L), Brad(L), Abby(W)

Erin(L): Erin(L), Brad(L), Dave(L), Abby(W)

Brad is now a Looser.

 

The next pair to be affirmed is Abby > Dave.  The Lists do not change.

 

The next pair to be affirmed is Brad > Abby.  

Abby(W): Abby(W), Brad(L), Dave(L), Erin(L)

Brad(L):Brad(L), Dave(L), Erin(L), Abby(W)

Cora(W): Cora(W)

Dave(L): Dave(L), Erin(L), Brad(L), Abby(W)

Erin(L): Erin(L), Brad(L), Abby(W), Dave(L)

Abby is still a Winner.

 

The next affirmed pair is Abby > Cora.  

Abby(W): Abby(W), Brad(L), Dave(L), Erin(L)

Brad(L):Brad(L), Dave(L), Erin(L), Abby(W)

Cora(L): Cora(L), Abby(W), Brad(L),
Dave(L), Erin(L)

Dave(L): Dave(L), Erin(L), Brad(L), Abby(W)

Erin(L): Erin(L), Brad(L), Abby(W), Dave(L)

Cora is now a looser.

 

Abby is the winner of the election.

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