[EM] What are some simple methods that accomplish the following conditions?

[Kristofer Munsterhjelm]

On 31/05/2019 00.33, Forest Simmons wrote:
> In the example profiles below 100 = P+Q+R, and  50>P>Q>R>0.  One
> consequence of these constraints is that in all three profiles below the
> cycle A>B>C>A will obtain.
> 
> I am interested in simple methods that always ...
> 
> (1) elect candidate A given the following profile:

[snip]

You might be able to do something with my (three-candidate) fpA-fpC
method, since it elects A in the situation where:

P: A
Q: B>C
R: C

since you have an ABCA cycle. In a 3-cycle, fpA-fpC lets each
candidate's score be the number of first preferences for that candidate,
minus the number of first preferences for whoever beats him pairwise.
Highest score wins. Thus the scores become:

A: fpA - fpC = P - R
B: fpB - fpA = Q - P
C: fpC - fpB = R - Q

Since P > Q > R, P - R > 0, but Q-P and R-Q < 0, so A wins.

Extending this method to four candidates so it meets Smith and still
both resists strategy and passes mono-raise is hard, and is one of the
things I'm working on (on and off) at the moment.