Thanks. Do you think that it is possible to make such an example without a sub-cycle? Because if not I am not overly perturbed. If you don't have a ready answer I'll get around to thinking more about it myself. We have the option of solving the clone problem by making it a one-at-time elimination method: use it find the lowest-ordered candidate, then eliminate that candidate and repeat and so on until one remains. (But of course that makes it more complex and possibly stuffs up mono-raise.) In your first matrix it was the diagonal row of "500"s that confused me. I was expecting it to be blanks or zeroes. Chris > *Joshua Boehme*joshua.p.boehme at gmail.com > <mailto:election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20%22Margin%20Sorted%20Minimum%20Losing%20Votes%20%28equal%20rated%20whole%29%22%0A%20candidate%20in%20poll&In-Reply-To=%3C0829ef7b-1d7f-4c1c-8d7a-8b19f9ec3f1b%40gmail.com%3E> > /Fri Apr 12 04:16:24 PDT 202*4*/ > ------------------------------------------------------------------------ > I manually tweaked a margin matrix from an existing example I keep on > hand. In case you find it easier to think in those terms (I do), here > it is... A B C D E A 0 50 -30 -30 -30 B -50 0 20 20 20 C 30 -20 0 100 > -300 D 30 -20 -100 0 200 E 30 -20 300 -200 0 The original emails > defined the method in terms of votes cast for each candidate, so I > restated it in those terms (and added enough votes to make it > feasible). With completely ordered ballots, I think it doesn't matter > which of those two bases you apply the method to. Here's one set of > ballots that produces those pairwise comparisons and has C, D, and E > as clones on every ballot: 35 ABCDE 450 BADEC 115 CDEAB 200 CEDAB 175 > ECDAB 25 BECDA On 4/12/24 05:05, Chris Benham wrote: >/Joshua, />//>/I > am finding this pairwise matrix confusing and hard to understand. > />//>/Is there any chance we can see the original ballot set? > />//>/Chris Benham/ > If I'm understanding this correctly, this method isn't always > cloneproof. Consider the following pairwise matrix (entries are row > candidate over column candidate)... A B C D E A 500 525 485 485 485 B > 475 500 510 510 510 C 515 490 500 550 350 D 515 490 450 500 600 E 515 > 490 650 400 500 The initial ordering is: A B D E C 485 475 450 400 350 > which is pairwise correct so we don't switch any candidates. If you > drop D and E, which are clones of C, you get: C A B 490 485 475 which > is also pairwise correct.

My non-rigorous guess is that it probably does require a sub-cycle, as that's the Achilles heel for a lot of methods that aren't strictly cloneproof. The concern for me is that it suggests clusters of similar candidates (not literal clones) can distort each others' results. For example, suppose voters first vote on left versus right, then rank within those two categories. On 4/12/24 23:32, Chris Benham wrote: > Thanks. Do you think that it is possible to make such an example without a sub-cycle? > > Because if not I am not overly perturbed. If you don't have a ready answer I'll get around to > thinking more about it myself. > > We have the option of solving the clone problem by making it a one-at-time elimination method: > use it find the lowest-ordered candidate, then eliminate that candidate and repeat and so on until > one remains. > > (But of course that makes it more complex and possibly stuffs up mono-raise.) > > In your first matrix it was the diagonal row of "500"s that confused me. I was expecting it to be blanks > or zeroes. > > Chris