---------- Forwarded message --------- From: Forest Simmons <fsimmons@pcc.edu> Date: Sat, Jun 1, 2019 at 12:01 PM Subject: Re: [EM] What are some simple methods that accomplish the following conditions? To: Kristofer Munsterhjelm <km_elmet@t-online.de> Great! This is the kind of creativity that will continue to keep this EM list alive and relevant. On Sat, Jun 1, 2019 at 8:13 AM Kristofer Munsterhjelm <km_elmet@t-online.de> wrote: > 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. >