[EM] are ranked pairs and river Schwartz-consistent?

[Jobst Heitzig]

Dear James!

I don't think you made a mistake with that example, only you forgot to
apply some sophisticated tiebreaker to the tie! Both ranked pairs and
river should use Steve's random voter hierarchy technique to resolve
that tie, so that essentially we get another non-majoritarian defeat A>R
or R>A with a smaller "strength" than all the majoritarian defeats.

When we always apply such a tiebreaker, it seems that the defeats always
build a "tournament" (=complete asymmetric binary relation), so the
Schwartz set is always the same as the Smith set or top-cycle, which
makes many things easier.

My best, Jobst


> Dear election methods fans,
> 
> 	I've often heard that ranked pairs is Schwartz consistent, but today I've
> been wrestling with an example where it doesn't seem to be. Hopefully I'm
> just making a mistake somewhere.
> 	First of all, by Schwartz, I mean in particular Schwartz's "union of
> minimal undominated sets" set, or GOCHA set.... defined as follows:
> 1. An undominated set is a set such that no candidate within the set is
> beaten by any candidate outside the set. 
> 2. A minimal undominated set does not contain any other undominated sets
> besides itself.
> 3. The GOCHA set is the union of minimal undominated sets.
> 
> Here is my example.
> Votes (with > symbol implied between each letter, e.g. 5 voters for
> R>S>A>T)
> 5: RSAT
> 5: TARS
> 4: STAR
> 4: RAST
> 3: TRAS
> 3: SATR
> 2: ATRS
> 2: SRAT
> 1: ASRT
> 1: TRAS
> 
> Pairwise comparisons
> A=R : 15-15
> A>S : 16-14
> A>T : 17-13
> R>S : 20-10
> S>T : 19-11
> T>R : 18-12
> 	As far as I can tell, the GOCHA set in this example is only {A}. A is
> surely undominated, and I cannot find any other sets of candidates other
> than {A B C D} which are also undominated. So, a Schwartz-efficient method
> should choose A with certainty.
> 	However, using ranked pairs, there is a tie between A and R, and if you
> use a basic tie-breaking ranking of the options method, there is a 50%
> chance that R will be elected.
> 	That is, using ranked pairs, the defeats are considered as follows
> 20-10		R>S		lock
> 19-11		S>T		lock
> 18-12		T>R		skip
> 17-13		A>T		lock
> 16-14		A>S		lock
> 	You are left with a tie, A=R>S>T. 
> 	The river method produces the same result.
> 20-10		R>S		lock
> 19-11		S>T		lock
> 18-12		T>R		skip
> 17-13		A>T		skip
> 16-14		A>S		skip
> 	A and R are both still undefeated.
> 	So, I don't understand what's going on. It doesn't seem like either
> ranked pairs or river choose A with certainty. 
> 	Again, I'm hoping that I have made a mistake somewhere, and there is
> someone on this list who can identify and correct it.
> 
> my best,
> James
> 
>