<div dir="auto">Chris,<div dir="auto"><br></div><div dir="auto">My suggestion in the replyI gave to your message (quoted below) made use of a "decloned Borda" score in the form of MaxPairwise Support plus MinPairwise Support, which sum is identical to Total Pairwise Support in the case of three candidate ... which... in turn, is the pairwise formulation of ordinary Borda.</div><div dir="auto"><br></div><div dir="auto">Additionally in the case of three candidates it is the same as your default approval.</div><div dir="auto"><br></div><div dir="auto">Even with many candidates MaxPS plus MimPS is the same as your default approval ... when we adopt the interpretatiomn ... that MinPS(X) is the pairwise support of the ballots fin favor of X over the virtual default approval cutoff candidate as a virtual candidate</div><div dir="auto">i.e.. .. the number of ballots on which Xi.e.anked below mobody.</div><div dir="auto"><br></div><div dir="auto">Similarly, if we count as a virtual candidate the bottom count cutoff, then MaxPS(X) is the same as the Implicit Approval of X.</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto">Then MaxPS plus MinPS is the zImlicit Approval plus the Top Count.</div><div dir="auto"><br></div><div dir="auto">So MinPlusMaxPS Is the same as Declomed Borda, which in turn, is the same as Default Approval.</div><div dir="auto"><br></div><div dir="auto">Then including the possibility of explicit approval, the method I proposed generalizes to ...</div><div dir="auto"><br></div><div dir="auto">For each candidate X, let Nemesis(X) be the most approved candidate that defeats X pairwise.</div><div dir="auto"><br></div><div dir="auto">And let diff(X) be the difference </div><div dir="auto"><br></div><div dir="auto">Approval(NemesisX) ) - Approval(X).</div><div dir="auto"><br></div><div dir="auto">Let L be the max diff((X) constrained by X having a two step beatpathback to to Nemesis(L).</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto">Elect W =Nemesis(L</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Jan 27, 2024, 7:51 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com">forest.simmons21@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><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" target="_blank" rel="noreferrer">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>
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