<div dir="auto">Great methods that avoid memtioning the controversial name of Condorcet <div dir="auto"><br></div><div dir="auto">Incorporating short beatpaths is an idea I.lke as well ... as in the following method:</div><div dir="auto"><br></div><div dir="auto">Elect the winner W of the greatest Borda margin defeat against any candidate L that has a two step beatpath back to any candidate that pairbeats it ... perhaps, for example, L beats X beats W.</div><div dir="auto"><br></div><div dir="auto">The motivation is that a buried beats-all candidate will always have a two step beattpath back to any candidate that directly defeated it via the very candidate X that expected to benefit from the burial.</div><div dir="auto"><br></div><div dir="auto">AND</div><div dir="auto"><br></div><div dir="auto">When one candidate is buried by another, the Borda margin between th the winner W and the loser L increases by lowering the Borda count of the buried candidate and raising the count of the "bus".that was raised to create the pairwise defeat in question.</div><div dir="auto"><br></div><div dir="auto">It is useful to know that the Borda Count for a candidate is the same as the sum of its pairwise supports ..the max plus te mim I'm the case of rwo matchups candidates ... which is a recipe for "declined Norda" in general ..in the match plus mim pairwise support.... which should be used in place of ordinary Borda where clone independence is desired.</div><div dir="auto"><br></div><div dir="auto">Note that the winner W will be a member of Smith because it will have a bestpath through L to any other candidate ...</div><div dir="auto"> even though no mention of Smth or Comdorcet was needed.</div><div dir="auto"><br></div><div dir="auto">It turns out empirically to be quite resistant to truncation offensives, as well.</div><div dir="auto"><br></div><div dir="auto">What do you think?</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Jan 27, 2024, 2:36 PM C.Benham <<a href="mailto:cbenham@adam.com.au">cbenham@adam.com.au</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><br>
I think Margins Sorted Top Ratings would be a good relatively burial <br>
resistant public proposal.<br>
<br>
* Voters rank from the top however many candidates they wish. <br>
Equal-ranking allowed.<br>
<br>
Give each candidate a score according to the number of ballots on which <br>
they are ranked below<br>
no others.<br>
<br>
Line them up in that order, highest to lowest.<br>
<br>
Check to see if all the candidates above bottom in this order pairwise <br>
beat the candidate immediately<br>
below them.<br>
<br>
If they do then elect the candidate highest in the order.<br>
<br>
If not begin with the pair that is pairwise out of order by the highest <br>
margin and swap them.<br>
(if there is an exact tie in the size of the margin then swap the <br>
tied-margin pair lowest in the order).<br>
<br>
Repeat until no pair of adjacent candidates are pairwise out of order <br>
and then elect the highest-ordered<br>
candidate. *<br>
<br>
This could also use ratings ballots.<br>
<br>
This meets Condorcet, but can be at least be explained (if not sold) <br>
without reference to Condorcet or Smith.<br>
<br>
It would be as monotonic as it is possible for a Condorcet method to be.<br>
<br>
For the sake of simplicity (and elegance) it has some short-comings. <br>
When there is a top cycle, voters who<br>
didn't top-rate (rank below no other candidates) any of the candidates <br>
in the Smith set are disadvantaged by comparison<br>
those that did. It would also fail Clone-Independence.<br>
<br>
A much more complicated method idea I had (that would be the same thing <br>
with three candidates):<br>
<br>
*Voters rank from the top however many candidates they wish. <br>
Equal-ranking allowed.<br>
<br>
(1) Eliminate (drop from the ballots and henceforth ignore) all <br>
candidates not in the Smith set.<br>
<br>
(2) Score the remaining candidates according to their minimum pairwise <br>
scores, with ballots that rank two candidates<br>
equal-top contributing a whole vote to each of the two candidate's <br>
scores against each other. Otherwise ballots that<br>
rank two candidates equal below top contribute zero to their pairwise <br>
scores against each other.<br>
<br>
(A possible variation is that they contribute half a vote to each if <br>
they are ranked below top and above bottom.)<br>
<br>
(3) Eliminate all candidates that don't have a "short" (one or two <br>
steps) beatpath to every candidate with a higher minimum<br>
pairwise score.<br>
<br>
(4). Repeat step 2. Then margins-sort the resulting scores and elect <br>
the highest-ordered candidate.*<br>
<br>
This is trying to meet Clone Independence, Mono-raise, Chicken Dilemma, <br>
Non-Drastic Defense.<br>
<br>
Chris Benham<br>
<br>
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</blockquote></div>