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

[Steve Eppley]

Hi,

James G-A wrote:
> 	I've often heard that ranked pairs is Schwartz consistent, 
-snip-

No, but it elects within the top cycle, which is nearly 
the same.  

> 	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.
-snip-

R tied A pairwise, which is fairly direct evidence that
society is split between R and A.  The majority that 
ranked another candidate over R is outweighed by the 
larger majorities in that cycle.  There's no majority
that would complain A should have been elected when 
R is elected.  So why worry?

I don't recall seeing any papers by Schwartz in which
he argued for Schwartz consistency over top cycle 
consistency.  Is there one?  If so, what was his 
justification?

--Steve