How to automatically convert ranked ballot sets into score ballot sets. This automatic conversion has been attempted in the past under the heading of Designated Strategy Voting (DSV) ... with mixed results ... not anything to write home about. A very crude attempt was Borda's rank scoring system, which is sometimes used in scoring team sports tournaments, for example. A new approach is based on two basic facts that are becoming more and more obvious: 1. It is easy to convert sets of ranked choice ballots to three-slot ballots in a monotonic, clone-free, decisive way. 2. There are many good monotonic, clone free, decisive ways of scoring three-slot ballots. In this message we will concentrate on the first fact, since it is the most recent major advancement in this topic. [Everybody and their dog has their favorite solution to the second of these two steps. I repeat the first step from the forwarded message that generically denominates the three respective slots Top, Middle, and Bottom, for want of better nomenclature: First generate a table of pairwise defeats to consult while making a second (final) pass through the input ballots. Then for each candidate k and each ballot B, decided whether B increments the Top, Bottom, or Mid level count of candidate k as follows: If on ballot B candidate k outranks some candidate j that defeats every candidate that outranks k, then increment the Top level count of candidate k. Else-If some candidate j ranked above k defeats every candidate that is outranked by k, then increment the Bottom level count of candidate k. Else increment the Mid level count of candidate k. End-If How has this simple procedure gone so long without discovery? The idea came to me when Rob Lanphier introduced me to "fear anchor" terminology. The candidate that defeats everybody you like better than k is an example of a fear anchor. On the other hand, the candidate that outranks k and defeats everybody outranked by k, could be called a "hope anchor." If k outranks some fear anchor, then k belongs in the Top slot. If k is out ranked by some hope anchor, then relegate k to the bottom slot. Is it possible for both?: if a hope anchor h outranks k, and a fear anchor f is outranked by k, then h defeats every candidate (such as f) that k outranks, and f defeats every candidate (such as h) that outranks k. So h defeats f, and f defeats h. Which means that the answer is no ... it is not possible for a ballot B to increment both Top and Bottom level counts for the same candidate k. Note that when there is a ballot CW, it is a fear anchor for every candidate ranked above it, and a hope anchor for every candidate that is outranks ... which means that candidates outrsnking the CW always get Top status, while candidates outranked by the CW get Bottom status. It easy to show that Top and Bottom candidates on the (pre-converted) ranked ballots preserve their status as such under the conversion to three-slot ballots. It is also easy to show that candidates awarded Top or Bottom status respectively by ballot B, form, respectively, Top and Bottom anchored solid coalitions on ballot B. What's not to like? -Forest ---------- Forwarded message --------- De: Forest Simmons <forest.simmons21@gmail.com> Date: lun., 2 de may. de 2022 12:47 p. m. Subject: Re: [EM] Definite Approval/Disapproval To: Rob Lanphier <roblan@gmail.com> While people are playing with election methods based on three level ballots, I would like to suggest another way to generate these three level ballots to reinforce my main contention... that three levels are easier to generate than two. This method transforms standard sets of Universal Domain style, ranked preference, ordinal ballots into 3 level ballots, which you may consider as cardinal, grade, or judgment category ballots as you so desire. A creative, accepting, brainstorming environment is what this thread is all about! First generate a table of pairwise defeats and ties to consult while making a second (final) pass through the input ballots. Then for each candidate k and each ballot B, decided whether B increments the Top, Bottom, or Mid level count of candidate k as follows: If on ballot B candidate k outranks some candidate j that is not defeated by any candidate ranked above k, then increment the Top level count of candidate k. ElseIf some candidate j ranked above k is not defeated by any candidate that is outranked by k, then increment the Bottom level count of candidate k. Else increment the Mid level count of candidate k. Isn't that much simpler than any (two level) approval DSV method that you've ever heard of? Of course, for three levels, Implicit Three Level would be even simpler: just count the ranked ballot Tops, Bottoms, and Middles without the possibility of dipping down or reaching up into the middle ranks for additional approvals or disapprovals. But where's the fun in that? Now keep grokking a good election method (besides simple Score and "Explicit Approval" ... the low hanging fruit already grabbed up my Rob Lanphier) based on three level ballots. Thanks. -Forest El dom., 1 de may. de 2022 8:00 p. m., Forest Simmons < forest.simmons21@gmail.com> escribió: > > > El sáb., 30 de abr. de 2022 8:58 p. m., Rob Lanphier <roblan@gmail.com> > escribió: > >> Hi Forest, >> >> Thanks for the food for thought (really.... I took a long time >> composing this email). My hunch is that you're proposing something >> awfully similar to "explicit approval" as devised by the folks in the >> Wikimedia community, but more inline below: >> >> On Sat, Apr 30, 2022 at 12:42 PM Forest Simmons >> <forest.simmons21@gmail.com> wrote: >> > On paper Approval has the best standard criteria compliances of any >> > method. On top of that it has all around simplicity going for it. Yet >> > nobody likes it, >> >> That's a peculiar definition of "nobody": >> >> https://electionscience.org/commentary-analysis/approval-voting-americas-favorite-voting-reform/ >> >> > and the most common complaint from experts and lay >> > citizens alike is that the definition of "approval" is so vague that >> > it could drive an indecisive person to distraction: There is no clear >> > guideline for partitioning the candidates into two distinct categories >> > with a crisp boundary between them. >> >> I agree with this complaint, and stated it as my objection for many >> years, but I've gotten over it. I'll restate a couple heuristics that >> I posted in a reddit comment recently: >> >> https://www.reddit.com/r/EndFPTP/comments/u1qguy/recordsetting_15_candidates_vie_for_fargo_city/i4eb3sw/ >> >> HEURISTIC A: >> 1. Find the candidate which seems likely to get elected, and that the >> voter is afraid will win. That's the "fear anchor" candidate. >> 2. For each candidate on the ballot, decide: >> 2a. if the candidate is better than the "fear anchor", then vote for them >> 2b. if the candidate is NOT better than the "fear anchor", then DO NOT >> vote for them >> >> HEURISTIC B: >> If the "frontrunner" is a good candidate, then vote for them. If not, >> then don't. Then decide on whether to vote for a candidate based on >> how they compare to the frontrunner (if better, then "YES"; if not >> better, then "NO"). >> > > Michael Ossipoff had a couple of other good ones: 1. Approve whomever you > would vote for in a FPTP Plurality election election as well as everyone > you like better. 2. Approve every candidate that you prefer over another > trip to the polls. > > Joe Weinstein: approve or disapprove X depending on whether you the winner > is more likely to be worse than X or better than X. > > Rob Legrand: Approve down to the most likely winner ... inclusive only if > the runner-up is worse. > > I don't to tip my hand yet ... I want people to grapple with it themselves. > > >> > In close second is the related complaint about lack of expressive power >> > ... in particular the inability to distinguish favorite, compromise, >> > and anti-favorite with three separate levels of ballot support. >> >> I think this is what the appeal of STAR voting is. But it's also what >> I liked about the form of explicit approval that Wikimedia Foundation >> used to use for many of its elections: >> https://electowiki.org/wiki/Explicit_approval_voting >> >> My hunch is that the Wikimedia folks got it right with their >> tabulation method. They had three levels for each candidate (Support >> / Abstain / Oppose) and then used the formula below to tabulate: >> >> Support / (Support + Oppose) >> >> ...and then relying on a per-candidate. Default was "abstain", and >> the quota was recalculated for each candidate. The system was biased >> against candidates that didn't elicit either strong support or strong >> opposition (since those candidates would have a difficult time meeting >> quota, since abstentions didn't count), but it seemed like a >> reasonable level of work to place on voters (to research candidates) >> and on candidates (to campaign, and increase their name recognition) >> >> Weirdly, English Wikipedia doesn't have an article for "Explicit >> approval voting", but it has "Combined approval voting": >> https://en.wikipedia.org/wiki/Combined_approval_voting >> >> ...which appears to select equivalent winners to Score voting with +0, >> +1, and +2 (integers) as the only options.. >> >> Regardless, are you proposing a fourth tabulation method? >> >> > "Definite Dis/Approval" (DD/A) addresses head on the basis for >> > these complaints with judgment style ballot instructions... "mark as >> > definitely approved (DA) only the candidates that you are absolutely >> > sure that you want to support" ... "mark as definitely disapproved >> > (DD) only the candidates that you very strongly feel to be unsuitable >> > for the position." Otherwise, mark the remaining candidates as >> > either ... "somewhere in the middle (Mid) between strongly suitable >> > and strongly unsuitable" or "No basis (NB) for an opinion." A blank >> > (i.e.abstemtion/undecided) is counted with the NB's. >> > >> > What if I'm not sure? Then most definitely it would be dishonest to >> > mark DA or DD. If you cannot decide between NB and Mid, then leave it >> > blank ... the ultimate expression of "Undecided". >> > >> > Now suppose that for each candidate k, you have the total counts DA(k), >> > DD(k), Mid(k), and NB(k), and no other information. >> > >> > 1. How would you use those totals to decide the single winner? >> > >> > 2. How would you construct a finish order if need be? >> > >> > 3. How would you resolve ties? >> >> Why come up with a new name and a new set of complicated jargon if you >> haven't answered these questions yet (especially since "Mid(k)" and >> "NB(k)" seem to be equivalent, and "NB" in my mind means "nota bene")? >> Explicit Approval, Combined Approval, and STAR voting answer your >> questions in three different ways, and all three of them have worthy >> cases for them (and against them). Perhaps a good starting point is >> to come up with your own answers to each of those questions, and then >> express your case using the language already used to describe one (or >> more) of those election methods. >> >> Sorry if the tone of my email seems negative. It just seems to me >> that much of the discussion on this mailing list is between people who >> want to invent their own jargon and their own election methods. I've >> been guilty of it myself (e.g. when several of us devised MATT and MAF >> in discussions on this mailing list back in 2018)[1][2][3]. Back in >> 2018 (after having spent many weekends the prior summer in California >> knocking on doors outside my district), and then one weekend tabling >> for the Center for Election Science, I came to realize that approval >> voting could solve problems in the California primaries, and tried to >> come up with a system that was simple enough, and didn't have the >> appearance of a difficult algebra problem photocopied from a linear >> algebra textbook. Can you (or someone on this list) come up with a >> system that's suitable for replacing the "blanket primary"[4] we have >> here in California? >> >> Rob >> [1]: >> http://lists.electorama.com/pipermail/election-methods-electorama.com//2018-November/thread.html >> [2]: >> http://lists.electorama.com/pipermail/election-methods-electorama.com//2018-December/thread.html >> [3]: https://electowiki.org/wiki/Approval-based_primary_election_methods >> [4]: https://electowiki.org/wiki/Blanket_primary >> >

I just realized that whenever the Smith set has three members, the member ranked between the other two is (by this new method) a cutoff between the two extremes. So in this case, the middle category "Mid" does not get used at all. In particular, all seven of Kevin's test cases are reduced to ordinary Approval by this DSV method. Are there any three slot examples with four member Smith sets for us to test? -Forest El mar., 3 de may. de 2022 2:47 p. m., Forest Simmons < forest.simmons21@gmail.com> escribió: > How to automatically convert ranked ballot sets into score ballot sets. > > This automatic conversion has been attempted in the past under the heading > of Designated Strategy Voting (DSV) ... with mixed results ... not anything > to write home about. > > A very crude attempt was Borda's rank scoring system, which is sometimes > used in scoring team sports tournaments, for example. > > A new approach is based on two basic facts that are becoming more and more > obvious: > > 1. It is easy to convert sets of ranked choice ballots to three-slot > ballots in a monotonic, clone-free, decisive way. > > 2. There are many good monotonic, clone free, decisive ways of scoring > three-slot ballots. > > In this message we will concentrate on the first fact, since it is the > most recent major advancement in this topic. [Everybody and their dog has > their favorite solution to the second of these two steps. > > I repeat the first step from the forwarded message that generically > denominates the three respective slots Top, Middle, and Bottom, for want of > better nomenclature: > > First generate a table of pairwise defeats to consult while making a > second (final) pass through the input ballots. > > Then for each candidate k and each ballot B, decided whether B increments > the Top, Bottom, or Mid level count of candidate k as follows: > > If on ballot B candidate k outranks some candidate j that defeats every > candidate that outranks k, then increment the Top level count of candidate > k. > > Else-If some candidate j ranked above k defeats every candidate that is > outranked by k, then increment the Bottom level count of candidate k. > > Else increment the Mid level count of candidate k. > > End-If > > How has this simple procedure gone so long without discovery? > > The idea came to me when Rob Lanphier introduced me to "fear anchor" > terminology. The candidate that defeats everybody you like better than k is > an example of a fear anchor. > > On the other hand, the candidate that outranks k and defeats everybody > outranked by k, could be called a "hope anchor." > > If k outranks some fear anchor, then k belongs in the Top slot. > > If k is out ranked by some hope anchor, then relegate k to the bottom slot. > > Is it possible for both?: if a hope anchor h outranks k, and a fear anchor > f is outranked by k, then h defeats every candidate (such as f) that k > outranks, and f defeats every candidate (such as h) that outranks k. > > So h defeats f, and f defeats h. > > Which means that the answer is no ... it is not possible for a ballot B to > increment both Top and Bottom level counts for the same candidate k. > > Note that when there is a ballot CW, it is a fear anchor for every > candidate ranked above it, and a hope anchor for every candidate that is > outranks ... which means that candidates outrsnking the CW always get Top > status, while candidates outranked by the CW get Bottom status. > > It easy to show that Top and Bottom candidates on the (pre-converted) > ranked ballots preserve their status as such under the conversion to > three-slot ballots. > > It is also easy to show that candidates awarded Top or Bottom status > respectively by ballot B, form, respectively, Top and Bottom anchored solid > coalitions on ballot B. > > What's not to like? > > -Forest > > > > > ---------- Forwarded message --------- > De: Forest Simmons <forest.simmons21@gmail.com> > Date: lun., 2 de may. de 2022 12:47 p. m. > Subject: Re: [EM] Definite Approval/Disapproval > To: Rob Lanphier <roblan@gmail.com> > > > While people are playing with election methods based on three level > ballots, I would like to suggest another way to generate these three level > ballots to reinforce my main contention... that three levels are easier to > generate than two. > > This method transforms standard sets of Universal Domain style, ranked > preference, ordinal ballots into 3 level ballots, which you may consider as > cardinal, grade, or judgment category ballots as you so desire. A creative, > accepting, brainstorming environment is what this thread is all about! > > First generate a table of pairwise defeats and ties to consult while > making a second (final) pass through the input ballots. > > Then for each candidate k and each ballot B, decided whether B increments > the Top, Bottom, or Mid level count of candidate k as follows: > > If on ballot B candidate k outranks some candidate j that is not defeated > by any candidate ranked above k, then increment the Top level count of > candidate k. > > ElseIf some candidate j ranked above k is not defeated by any candidate > that is outranked by k, then increment the Bottom level count of candidate > k. > > Else increment the Mid level count of candidate k. > > Isn't that much simpler than any (two level) approval DSV method that > you've ever heard of? > > Of course, for three levels, Implicit Three Level would be even simpler: > just count the ranked ballot Tops, Bottoms, and Middles without the > possibility of dipping down or reaching up into the middle ranks for > additional approvals or disapprovals. But where's the fun in that? > > Now keep grokking a good election method (besides simple Score and > "Explicit Approval" ... the low hanging fruit already grabbed up my Rob > Lanphier) based on three level ballots. > > Thanks. > > -Forest > > > > > El dom., 1 de may. de 2022 8:00 p. m., Forest Simmons < > forest.simmons21@gmail.com> escribió: > >> >> >> El sáb., 30 de abr. de 2022 8:58 p. m., Rob Lanphier <roblan@gmail.com> >> escribió: >> >>> Hi Forest, >>> >>> Thanks for the food for thought (really.... I took a long time >>> composing this email). My hunch is that you're proposing something >>> awfully similar to "explicit approval" as devised by the folks in the >>> Wikimedia community, but more inline below: >>> >>> On Sat, Apr 30, 2022 at 12:42 PM Forest Simmons >>> <forest.simmons21@gmail.com> wrote: >>> > On paper Approval has the best standard criteria compliances of any >>> > method. On top of that it has all around simplicity going for it. Yet >>> > nobody likes it, >>> >>> That's a peculiar definition of "nobody": >>> >>> https://electionscience.org/commentary-analysis/approval-voting-americas-favorite-voting-reform/ >>> >>> > and the most common complaint from experts and lay >>> > citizens alike is that the definition of "approval" is so vague that >>> > it could drive an indecisive person to distraction: There is no clear >>> > guideline for partitioning the candidates into two distinct categories >>> > with a crisp boundary between them. >>> >>> I agree with this complaint, and stated it as my objection for many >>> years, but I've gotten over it. I'll restate a couple heuristics that >>> I posted in a reddit comment recently: >>> >>> https://www.reddit.com/r/EndFPTP/comments/u1qguy/recordsetting_15_candidates_vie_for_fargo_city/i4eb3sw/ >>> >>> HEURISTIC A: >>> 1. Find the candidate which seems likely to get elected, and that the >>> voter is afraid will win. That's the "fear anchor" candidate. >>> 2. For each candidate on the ballot, decide: >>> 2a. if the candidate is better than the "fear anchor", then vote for them >>> 2b. if the candidate is NOT better than the "fear anchor", then DO NOT >>> vote for them >>> >>> HEURISTIC B: >>> If the "frontrunner" is a good candidate, then vote for them. If not, >>> then don't. Then decide on whether to vote for a candidate based on >>> how they compare to the frontrunner (if better, then "YES"; if not >>> better, then "NO"). >>> >> >> Michael Ossipoff had a couple of other good ones: 1. Approve whomever you >> would vote for in a FPTP Plurality election election as well as everyone >> you like better. 2. Approve every candidate that you prefer over another >> trip to the polls. >> >> Joe Weinstein: approve or disapprove X depending on whether you the >> winner is more likely to be worse than X or better than X. >> >> Rob Legrand: Approve down to the most likely winner ... inclusive only if >> the runner-up is worse. >> >> I don't to tip my hand yet ... I want people to grapple with it >> themselves. >> >> >>> > In close second is the related complaint about lack of expressive power >>> > ... in particular the inability to distinguish favorite, compromise, >>> > and anti-favorite with three separate levels of ballot support. >>> >>> I think this is what the appeal of STAR voting is. But it's also what >>> I liked about the form of explicit approval that Wikimedia Foundation >>> used to use for many of its elections: >>> https://electowiki.org/wiki/Explicit_approval_voting >>> >>> My hunch is that the Wikimedia folks got it right with their >>> tabulation method. They had three levels for each candidate (Support >>> / Abstain / Oppose) and then used the formula below to tabulate: >>> >>> Support / (Support + Oppose) >>> >>> ...and then relying on a per-candidate. Default was "abstain", and >>> the quota was recalculated for each candidate. The system was biased >>> against candidates that didn't elicit either strong support or strong >>> opposition (since those candidates would have a difficult time meeting >>> quota, since abstentions didn't count), but it seemed like a >>> reasonable level of work to place on voters (to research candidates) >>> and on candidates (to campaign, and increase their name recognition) >>> >>> Weirdly, English Wikipedia doesn't have an article for "Explicit >>> approval voting", but it has "Combined approval voting": >>> https://en.wikipedia.org/wiki/Combined_approval_voting >>> >>> ...which appears to select equivalent winners to Score voting with +0, >>> +1, and +2 (integers) as the only options.. >>> >>> Regardless, are you proposing a fourth tabulation method? >>> >>> > "Definite Dis/Approval" (DD/A) addresses head on the basis for >>> > these complaints with judgment style ballot instructions... "mark as >>> > definitely approved (DA) only the candidates that you are absolutely >>> > sure that you want to support" ... "mark as definitely disapproved >>> > (DD) only the candidates that you very strongly feel to be unsuitable >>> > for the position." Otherwise, mark the remaining candidates as >>> > either ... "somewhere in the middle (Mid) between strongly suitable >>> > and strongly unsuitable" or "No basis (NB) for an opinion." A blank >>> > (i.e.abstemtion/undecided) is counted with the NB's. >>> > >>> > What if I'm not sure? Then most definitely it would be dishonest to >>> > mark DA or DD. If you cannot decide between NB and Mid, then leave it >>> > blank ... the ultimate expression of "Undecided". >>> > >>> > Now suppose that for each candidate k, you have the total counts DA(k), >>> > DD(k), Mid(k), and NB(k), and no other information. >>> > >>> > 1. How would you use those totals to decide the single winner? >>> > >>> > 2. How would you construct a finish order if need be? >>> > >>> > 3. How would you resolve ties? >>> >>> Why come up with a new name and a new set of complicated jargon if you >>> haven't answered these questions yet (especially since "Mid(k)" and >>> "NB(k)" seem to be equivalent, and "NB" in my mind means "nota bene")? >>> Explicit Approval, Combined Approval, and STAR voting answer your >>> questions in three different ways, and all three of them have worthy >>> cases for them (and against them). Perhaps a good starting point is >>> to come up with your own answers to each of those questions, and then >>> express your case using the language already used to describe one (or >>> more) of those election methods. >>> >>> Sorry if the tone of my email seems negative. It just seems to me >>> that much of the discussion on this mailing list is between people who >>> want to invent their own jargon and their own election methods. I've >>> been guilty of it myself (e.g. when several of us devised MATT and MAF >>> in discussions on this mailing list back in 2018)[1][2][3]. Back in >>> 2018 (after having spent many weekends the prior summer in California >>> knocking on doors outside my district), and then one weekend tabling >>> for the Center for Election Science, I came to realize that approval >>> voting could solve problems in the California primaries, and tried to >>> come up with a system that was simple enough, and didn't have the >>> appearance of a difficult algebra problem photocopied from a linear >>> algebra textbook. Can you (or someone on this list) come up with a >>> system that's suitable for replacing the "blanket primary"[4] we have >>> here in California? >>> >>> Rob >>> [1]: >>> http://lists.electorama.com/pipermail/election-methods-electorama.com//2018-November/thread.html >>> [2]: >>> http://lists.electorama.com/pipermail/election-methods-electorama.com//2018-December/thread.html >>> [3]: https://electowiki.org/wiki/Approval-based_primary_election_methods >>> [4]: https://electowiki.org/wiki/Blanket_primary >>> >>

Hi Forest, you could start with a 1v1 and toss in noise candidates who can beat only one of them: 26 D>A 23 A 27 C>B 24 B Kevin (end) Le mercredi 4 mai 2022, 11:33:38 UTC−5, Forest Simmons <forest.simmons21@gmail.com> a écrit : I just realized that whenever the Smith set has three members, the member ranked between the other two is (by this new method) a cutoff between the two extremes. So in this case, the middle category "Mid" does not get used at all. In particular, all seven of Kevin's test cases are reduced to ordinary Approval by this DSV method. Are there any three slot examples with four member Smith sets for us to test? -Forest El mar., 3 de may. de 2022 2:47 p. m., Forest Simmons <forest.simmons21@gmail.com> escribió: > How to automatically convert ranked ballot sets into score ballot sets. > > This automatic conversion has been attempted in the past under the heading of Designated Strategy Voting (DSV) ... with mixed results ... not anything to write home about. > > A very crude attempt was Borda's rank scoring system, which is sometimes used in scoring team sports tournaments, for example. > > A new approach is based on two basic facts that are becoming more and more obvious:  > > 1. It is easy to convert sets of ranked choice ballots to three-slot ballots in a monotonic, clone-free, decisive way. > > 2. There are many good monotonic, clone free, decisive ways of scoring three-slot ballots. > > In this message we will concentrate on the first fact, since it is the most recent major advancement in this topic. [Everybody and their dog has their favorite solution to the second of these two steps. > > I repeat the first step from the forwarded message that generically denominates the three respective slots Top, Middle, and Bottom, for want of better nomenclature:  > > First generate a table of pairwise defeats to consult  while making a second (final) pass through the input ballots. > > Then for each candidate k and each ballot B, decided whether B increments the Top, Bottom, or Mid level count of candidate k as follows: > > If on ballot B candidate k outranks some candidate j that defeats every candidate that outranks k, then increment the Top level count of candidate k. > > Else-If some candidate j ranked above k defeats every candidate that is outranked by k, then increment the Bottom level count of candidate k. > > Else increment the Mid level count of candidate k. > > End-If  > > How has this simple procedure gone so long without discovery? > > The idea came to me when Rob Lanphier introduced me to "fear anchor" terminology. The candidate that defeats everybody you like better than k is an example of a fear anchor. > > On the other hand, the candidate that outranks k and defeats everybody outranked by k, could be called a "hope anchor." > > If k  outranks some fear anchor, then k belongs in the Top slot. > > If k is out ranked by some hope anchor, then relegate k to the bottom slot. > > Is it possible for both?: if a hope anchor h outranks k, and a fear anchor f is outranked by k, then h defeats every candidate (such as f) that k outranks, and f defeats every candidate (such as h) that outranks k. > > So h defeats f, and f defeats h.  > > Which means that the answer is no ... it is not possible for a ballot B to increment both Top and Bottom level counts for the same candidate k. > > Note that when there is a ballot CW,  it is a fear anchor for every candidate ranked above it, and a hope anchor for every candidate that is outranks ... which means that candidates outrsnking the CW always get Top status, while candidates outranked by the CW get Bottom status. > > It easy to show that Top and Bottom candidates on the (pre-converted) ranked ballots preserve their status as such under the conversion to three-slot ballots. > > It is also easy to show that candidates awarded Top or Bottom status respectively by ballot B, form, respectively, Top and Bottom anchored solid coalitions on ballot B. > > What's not to like? > > -Forest

Great! El mié., 4 de may. de 2022 1:16 p. m., Kevin Venzke <stepjak@yahoo.fr> escribió: > Hi Forest, you could start with a 1v1 and toss in noise candidates who can > beat only one of them: > 26 D>A > A put into Top with D because C beats D above A from below A. > 23 A > 27 C>B > B is put into top because A beats C from below B. > 24 B > So all implicit approval candidates get Top slot status in the conversion. The respective Top slot counts for A,B,C,D are 49, 51, 27, and 26. The Bottom Counts are 51, 49, 73, and 74, respectively. All Mid slot counts are zero. Most reasonable three slot methods would elect B. > [And most reasonable Condorcet methods would elect B from the original ballots as well: B beats A 51 to 49. B beats D 51 to 26. D beats A 26 to 23. A beats C 49 to 27. C beats B 27 to 24. C beats D 27 to 26.] > > Kevin > > (end) > Somebody with patience should go through the seven test examples to see how this conversion to 3-slots compares with other methods ... assuming approval/score as the 3-slot method to be used on the converted ballots. -Forest > > > Le mercredi 4 mai 2022, 11:33:38 UTC−5, Forest Simmons < > forest.simmons21@gmail.com> a écrit : > I just realized that whenever the Smith set has three members, the member > ranked between the other two is (by this new method) a cutoff between the > two extremes. So in this case, the middle category "Mid" does not get used > at all. > > In particular, all seven of Kevin's test cases are reduced to ordinary > Approval by this DSV method. > > Are there any three slot examples with four member Smith sets for us to > test? > > -Forest > > El mar., 3 de may. de 2022 2:47 p. m., Forest Simmons < > forest.simmons21@gmail.com> escribió: > > How to automatically convert ranked ballot sets into score ballot sets. > > > > This automatic conversion has been attempted in the past under the > heading of Designated Strategy Voting (DSV) ... with mixed results ... not > anything to write home about. > > > > A very crude attempt was Borda's rank scoring system, which is sometimes > used in scoring team sports tournaments, for example. > > > > A new approach is based on two basic facts that are becoming more and > more obvious: > > > > 1. It is easy to convert sets of ranked choice ballots to three-slot > ballots in a monotonic, clone-free, decisive way. > > > > 2. There are many good monotonic, clone free, decisive ways of scoring > three-slot ballots. > > > > In this message we will concentrate on the first fact, since it is the > most recent major advancement in this topic. [Everybody and their dog has > their favorite solution to the second of these two steps. > > > > I repeat the first step from the forwarded message that generically > denominates the three respective slots Top, Middle, and Bottom, for want of > better nomenclature: > > > > First generate a table of pairwise defeats to consult while making a > second (final) pass through the input ballots. > > > > Then for each candidate k and each ballot B, decided whether B > increments the Top, Bottom, or Mid level count of candidate k as follows: > > > > If on ballot B candidate k outranks some candidate j that defeats every > candidate that outranks k, then increment the Top level count of candidate > k. > > > > Else-If some candidate j ranked above k defeats every candidate that is > outranked by k, then increment the Bottom level count of candidate k. > > > > Else increment the Mid level count of candidate k. > > > > End-If > > > > How has this simple procedure gone so long without discovery? > > > > The idea came to me when Rob Lanphier introduced me to "fear anchor" > terminology. The candidate that defeats everybody you like better than k is > an example of a fear anchor. > > > > On the other hand, the candidate that outranks k and defeats everybody > outranked by k, could be called a "hope anchor." > > > > If k outranks some fear anchor, then k belongs in the Top slot. > > > > If k is out ranked by some hope anchor, then relegate k to the bottom > slot. > > > > Is it possible for both?: if a hope anchor h outranks k, and a fear > anchor f is outranked by k, then h defeats every candidate (such as f) that > k outranks, and f defeats every candidate (such as h) that outranks k. > > > > So h defeats f, and f defeats h. > > > > Which means that the answer is no ... it is not possible for a ballot B > to increment both Top and Bottom level counts for the same candidate k. > > > > Note that when there is a ballot CW, it is a fear anchor for every > candidate ranked above it, and a hope anchor for every candidate that is > outranks ... which means that candidates outrsnking the CW always get Top > status, while candidates outranked by the CW get Bottom status. > > > > It easy to show that Top and Bottom candidates on the (pre-converted) > ranked ballots preserve their status as such under the conversion to > three-slot ballots. > > > > It is also easy to show that candidates awarded Top or Bottom status > respectively by ballot B, form, respectively, Top and Bottom anchored solid > coalitions on ballot B. > > > > What's not to like? > > > > -Forest >