Let's say that a subset S of candidates is "uninterrupted" on ballot B iff no candidate that is not a member of S is ranked between two members of S. An uninterrupted set that contains some top ranked candidate is "top tethered." Similarly, a "bottom tethered" uninterrupted set has at least one candidate that does not outrank any candidate. Every "solid coalition" of Woodall is an example of a top tethered uninterrupted set. Each completely ranked ballot of n candidates has n of these (non-empty) solid coalitions, as also n bottom tethered uninterrupted sets, and many more untethered uninterrupted sets. In fact, the total number of uninterrupted sets on a fully ranked ballot of n candidates would have to be C(n+1,2) or n(n+1)/2, since it takes two cutoffs to delineate an uninterrupted set, and there are n+1 slots for those boundary marks. Let beta be a set of ballots. Then for each subset S of candidates, let UI(S) be the number of ballots in beta on which S is uninterrupted. For each ballot B we determine a representative candidate K(B) by considering the uninterrupted sets S in order of decreasing UI(S). When a set S is considered, every candidate not in the set becomes ineligible to represent ballot B, unless this would cause all candidates to be ineligible, in which case that set is ignored. When only one candidate is still eligible to represent ballot B, that candidate is selected as K(B). For each candidate k, let N(k) be the number of ballots for which k=K(B). De-cloned Copeland: Elect the candidate X with the greatest sum (over those k that do not defeat X) of N(k). Does that work? Really, there are at least three versions ... top tethered, bottom tethered, and untethered ... not to mention acquiescing variants. If the descending uninterrupted sets end in a tied set of candidates T(B) to represent ballot B, then B contributes to each of their N(k) values 1/#T(B). Is this the right way to adapt Woodall's idea for this context? -Forest

Hi Forest, Le mardi 10 mai 2022, 00:02:19 UTC−5, Forest Simmons <forest.simmons21@gmail.com> a écrit : > Let's say that a subset S of candidates is "uninterrupted" on ballot B  iff no candidate > that is not a member of S is ranked between two members of S. >  > An uninterrupted set that contains some top ranked candidate is "top tethered." Similarly, > a "bottom tethered"  > uninterrupted set has at least one candidate that does not outrank any candidate. >  > Every "solid coalition" of Woodall is an example of a top tethered uninterrupted set. > Each completely ranked ballot of n candidates has n of these (non-empty) solid coalitions, > as also n bottom tethered uninterrupted sets, and many more untethered uninterrupted sets. > In fact, the total number of uninterrupted sets on a fully ranked ballot of n candidates > would have to be C(n+1,2) or n(n+1)/2, since it takes two cutoffs to delineate an > uninterrupted set, and there are n+1 slots for those boundary marks. >  > Let beta be a set of ballots. Then for each subset S of candidates, let UI(S) be the number > of ballots in beta on which S is uninterrupted. This sounds like UI(S) measures how likely it is that S consists of clones. It seems like every singleton set would have a 100% score. (With the untethered definition.) > For each ballot B we determine a representative candidate K(B) by considering the > uninterrupted sets S in order of decreasing UI(S). > When a set S is considered, every candidate not in the set becomes ineligible to represent > ballot B, unless this would cause all candidates to be ineligible, in which case that set > is ignored. > When only one candidate is still eligible to represent ballot B, that candidate is selected > as K(B). If I'm right about the above then every ballot immediately runs into a massive tie. But supposing it doesn't, I am not sure why a high UI score suggests that the candidate should be representative of the ballot. > For each candidate k, let N(k) be the number of ballots for which k=K(B).  > De-cloned Copeland: > Elect the candidate X  with the greatest sum (over those k that do not defeat X) of N(k). > Does that work? >  > Really, there are at least three versions ... top tethered, bottom tethered, and untethered > ... not to mention acquiescing variants. I think your definition is already acquiescing. With top tethering you might cut down on the UI scores so that only clone sets at the tops of ballots register. Especially if you use solid and not acquiescing. The result could be that lone candidates have the highest UI score, according to their first preference count. It's a little unclear to me whether the choice of tethering rule affects the calculation of UI, the sets on B that can be considered, or both. If both, then I guess you'll usually end up with the first preference as K(B). > If the descending uninterrupted sets end in a tied set of candidates T(B) to represent > ballot B, then B contributes to each of their N(k) values 1/#T(B). This is potentially tedious because the possible results of all ties can't usually be found by a single pass through the sets (unless no sets have tied strength). And you need to assess this potentially for each unique ballot. > Is this the right way to adapt Woodall's idea for this context? Woodall used this concept in a lot of different ways (not just solid and acquiescing), including reproducing versions of Schulze. But in all cases the score of a set would correspond in some way to voter support for the set. Of course, he was always looking to find the winning candidate at the end. Maybe other applications are possible. Kevin

El mar., 10 de may. de 2022 5:28 p. m., Kevin Venzke <stepjak@yahoo.fr> escribió: > Hi Forest, > > Le mardi 10 mai 2022, 00:02:19 UTC−5, Forest Simmons < > forest.simmons21@gmail.com> a écrit : > > Let's say that a subset S of candidates is "uninterrupted" on ballot B > iff no candidate > > that is not a member of S is ranked between two members of S. > > > > An uninterrupted set that contains some top ranked candidate is "top > tethered." Similarly, > > a "bottom tethered" > > uninterrupted set has at least one candidate that does not outrank any > candidate. > > > > Every "solid coalition" of Woodall is an example of a top tethered > uninterrupted set. > > Each completely ranked ballot of n candidates has n of these (non-empty) > solid coalitions, > > as also n bottom tethered uninterrupted sets, and many more untethered > uninterrupted sets. > > In fact, the total number of uninterrupted sets on a fully ranked ballot > of n candidates > > would have to be C(n+1,2) or n(n+1)/2, since it takes two cutoffs to > delineate an > > uninterrupted set, and there are n+1 slots for those boundary marks. > > > > Let beta be a set of ballots. Then for each subset S of candidates, let > UI(S) be the number > > of ballots in beta on which S is uninterrupted. > > This sounds like UI(S) measures how likely it is that S consists of > clones. It > seems like every singleton set would have a 100% score. (With the > untethered > definition.) > I made the mistake of copying and pasting from the DSC article ... for the untethered you have to consider the sets in descending order of cardinality, and within cardinality, descending UI score. > > > For each ballot B we determine a representative candidate K(B) by > considering the > > uninterrupted sets S in order of decreasing UI(S). > > When a set S is considered, every candidate not in the set becomes > ineligible to represent > > ballot B, unless this would cause all candidates to be ineligible, in > which case that set > > is ignored. > > When only one candidate is still eligible to represent ballot B, that > candidate is selected > > as K(B). > > If I'm right about the above then every ballot immediately runs into a > massive > tie. But supposing it doesn't, I am not sure why a high UI score suggests > that the > candidate should be representative of the ballot. > The ballot's representative for the clone count. We are trying to get a probability measure (by normalizing the N(k) counts) on the set of candidates. We want the measure to nrespect clone sets and be neutral. The clone sets should come out identically for reversed ballots ... so also the probability measure. If a candidate is cloned the probability of original candidate should be distributed among its clones. It was a big mistake years ago for me to assume that the probabilities should be winning probabilities. > > For each candidate k, let N(k) be the number of ballotntstives for which > k=K(B). > > De-cloned Copeland: > > Elect the candidate X with the greatest sum (over those k that do not > defeat X) of N(k). > > Does that work? > > > > Really, there are at least three versions ... top tethered, bottom > tethered, and untethered > > ... not to mention acquiescing variants. > > I think your definition is already acquiescing. > > With top tethering you might cut down on the UI scores so that only clone > sets at > the tops of ballots register. Especially if you use solid and not > acquiescing. The > result could be that lone candidates have the highest UI score, according > to their > first preference count. > > It's a little unclear to me whether the choice of tethering rule affects > the > calculation of UI, the sets on B that can be considered, or both. If both, > then I > guess you'll usually end up with the first preference as K(B). > > > If the descending uninterrupted sets end in a tied set of candidates > T(B) to represent > > ballot B, then B contributes to each of their N(k) values 1/#T(B). > > This is potentially tedious because the possible results of all ties can't > usually > be found by a single pass through the sets (unless no sets have tied > strength). And > you need to assess this potentially for each unique ballot. > > > Is this the right way to adapt Woodall's idea for this context? > > Woodall used this concept in a lot of different ways (not just solid and > acquiescing), including reproducing versions of Schulze. But in all cases > the score > of a set would correspond in some way to voter support for the set. Of > course, he > was always looking to find the winning candidate at the end. Maybe other > applications are possible. > > Kevin >