On 08.11.2022 18:02, Richard, the VoteFair guy wrote:

 Or is

 but I

 guess.

 The short answer is: A covers B if A pairwise beats everybody B
 pairwise
 beats and then some.

 An uncovered candidate is someone who is not covered by anyone else.

 This definition works when there are no pairwise ties. Things get
 trickier with pairwise ties, as I found out when generalizing

 Cover.

 -km
 ----
 Election-Methods mailing list - see https://electorama.com/em
 <https://electorama.com/em> for list info

On Thu, Nov 10, 2022, 9:40 PM Richard, the VoteFair guy < electionmethods@votefair.org> wrote: > On 11/9/2022 12:05 AM, Forest Simmons wrote: > > > A candidate is uncovered iff it has a beatpath of > > only two steps to each candidate (if any) that beats it. > > Should the "only two steps" be interpreted as "two or fewer steps?" or > "two or more steps?" or "exactly two steps?" > In this context, exactly two, because when a candidate beats it, it cannot beat that candidate in one step, since it is not possible for X to outrank Y on more ballots than Y outranks X, and at the same time have Y out rank X on more ballots than X outranks Y. A candidate X has a beatpath to every other candidate if and only if X is in the Smith set. So the Landau is a subset of the Smith set. You see Landau is more special because it requires shorter beatpaths. Any length beatpath will do for Smith. Suppose the cycle ABCA is augmented with a candidate D that beats only A, and is beaten by B and C. Then all four candidates are in Smith because the beat cycle DABCD includes a beatpath from each candidate to each of the other candidates. Is D a Landau candidate? Well C beats it. So does it have a two-step path back to C? No. So it is covered by C, and covered means not in Landau. Is A a Landau candidate? Yes. Only D and C beat it, and ABD and ABC are the respective required two-step beatpaths that get back to D and C, respectively. Does that help? > I'm still trying to understand the "uncovered" (and "covered") concept. > Especially, what is its relationship to the Smith set? > > Thanks, > > Richard Fobes > > > On 11/9/2022 12:05 AM, Forest Simmons wrote: > > In this context the most relevant question is what do we mean by > > "uncovered", since that's the word used in the method definition ... > > > > Repeatedly eliminate the (remaining) candidate with fewest votes until > > there remains only one uncovered candidate to elect. > > > > No need to know what covering means, although you can figure it out > > indirectly from the definition of "uncovered:" > > > > A candidate is uncovered iff it has a beatpath of only two steps to each > > candidate (if any) that beats it. > > > > Any candidate X who complains that they should have won because they > > beat the winner W pairwise will get this truthful and obviously relevant > > rejoinder: > > > > When you were eliminated, you had fewer transferred votes than I. > > > > I fact, I beat every candidate pairwise that was not already eliminated > > (like you) on the basis of two few (transferred) votes. > > > > It is very easy to discern if some candidate X is uncovered: > > > > Just check each candidate Y that beats it (X) to see if it has a two > > step beatpath via some Z, back to Y: > > > > X beats Z beats Y > > > > Only Smith candidates can be uncovered because only Smith candidates > > have beatpaths back to the candidates that beat them. So the candidates > > you have to check are the Smith candidates ... at most three, and rarely > > more than one, in a public election. > > > > If you want, you can run IRV all the way through ... then if the IRV > > winner is uncovered, you are done. If not, back up until you cone to an > > uncovered candidate ... that's your winner! > > > > It's just a matter of doing regular IRV, and backing up (if necessary) > > until you get to an uncovered candidate. > > > > Forest > > > > > > On Tue, Nov 8, 2022, 11:18 AM Kristofer Munsterhjelm > > <km_elmet@t-online.de <mailto:km_elmet@t-online.de>> wrote: > > > > On 08.11.2022 18:02, Richard, the VoteFair guy wrote: > > > Forest, what do you mean by "covered"? Is there a Wikipedia or > > > Electowiki article (or section of an article) that explains it? > > Or is > > > there a dictionary reference you can point to? > > > > > > Yes, you've used the words "covered" and "uncovered" many times > > but I > > > don't recall ever seeing a clear explanation of what you mean. I > > > presume it involves pairwise counts, but that's as far as I can > > guess. > > > > The short answer is: A covers B if A pairwise beats everybody B > > pairwise > > beats and then some. > > > > An uncovered candidate is someone who is not covered by anyone else. > > > > This definition works when there are no pairwise ties. Things get > > trickier with pairwise ties, as I found out when generalizing > Friendly > > Cover. > > > > -km > > ---- > > Election-Methods mailing list - see https://electorama.com/em > > <https://electorama.com/em> for list info > > > ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

Grammatical correction... On Sat, Nov 12, 2022, 6:28 PM Forest Simmons <forest.simmons21@gmail.com> wrote: > I tried to state the most understandable description possible of Gross > Loser Elimination ... is the following version an improvement? Anybody have > a simpler or clearer description? > > Suppose there are seven candidates. Each candidate gets six cards with > their name at the top and one of the other candidate names just below it. > > Each card is given to a different election worker. > > Pretend you are in charge of the card with Diane at the top and Jenny just > below it. > > As the ballots are slowly opened one by one in a public ceremony so that > everybody can see each ballot on an overhead projector screen, you put a > hash mark on your Diane/Jenny card every time a ballot is opened that shows > a vote for Diane over Jenny. > > After the ballots have been tallied in this way, you total the votes, and > put that vote total to the left of her name on the top left of your card. > This is the total of her votes in her matchup against one of the other > candidates, namely Jenny. > > Then you hand in your card to the head vote counter, who sorts the cards > in order of the vote totals in the upper left corner of each card. > > Now the Elimination steps: > > Remove the card from the bottom of the stack, the one with the smallest > vote total next to the name at the top of that card. That name identifies > the Gross Loser. Eliminate all of the cards > that have that name anywhere on it. > Should be "on them" not "on it" That concludes one step of GLE, Gross Loser Elimination. > > Now remove the card from the bottom of the remaining deck. The name at > the top of that card is the Gross Loser name for this step.. Eliminate all > of the cards that have that name anywhere on it. That concludes another > step of GLE, Gross Loser Elimination. > > Now remove the card from the bottom of the remaining deck. The name at > the top of that card is the Gross Loser name for this step.. Eliminate all > of the cards that have that name anywhere on it. That concludes another > step of GLE, Gross Loser Elimination. > > Continue in this manner until all cards but two have been eliminated. > Eliminate the bottom of these two, and elect the candidate whose name is at > the top of the remaining card. This is the only candidate that was never at > any stage the worst of the worst ... i.e. never the Gross Loser. > > -Forest > > On Sat, Nov 12, 2022, 12:13 PM Forest Simmons <forest.simmons21@gmail.com> > wrote: > >> Q&D was the simplest way to always get the same result (when the Smith >> set was four or fewer members) as Implicit Approval Chain Climbing ... a >> monotonic, clone free, burial resistant, Banks efficient method ... as >> simple as possible for a method with those criteria compliances .... >> compliances that no other method on record could truthfully claim. >> >> So why did it get no traction? >> >> According guys "in the trenches" it has to be an elimination method with >> vote transfers between steps. >> >> No such method is monotonic, but the next best thing is Yee/Bolson >> monotonic. >> >> That method is Gross Loser Elimination (GLE). >> >> The Gross Loser of a Round Robin tournament is the player whose worst >> score (for any of her matchups) is worse than anybody elses's ... we could >> call her the MinMin loser. >> >> GLM is the elimination method that at each stage eliminates the gross >> loser of the remaing candidates ... after the gross losers from the >> previous stages have already been removed. The last candidate standing is >> the winner. >> >> As a reminder, the gross loser is the candidate with the worst worst >> score. Each candidate has several scores ... one against each of the other >> players. The worst of these is that candidate's worst score. The candidate >> whose worst score is worse than anybody elses's worst score is the gross >> loser. That is the one to be eliminated in the first step. >> >> In the second step each candidate has a worst score in any of its >> matchups with the remaining candidates. The candidate whose worst score is >> worse than any other remaining candidate's worst score is the gross loser >> of that stage. That's the one to be eliminated at that stage. >> >> For example suppose that in 10th stage there are only four candidates >> left ... candidates A B,C,&D with respective worst matchup scores of 15, >> 13, 28, and 35. Which one do you think would be eliminated at that stage? >> >> If you guessed candidate B, you guessed right: B's worst score (13) at >> this stage is worse (smaller than) any of the other three scores 15, 28, or >> 35. >> >> Each elimination step is that simple! >> >> Where do the matchup scores come from? In a sports tournament it is >> obvious because each matchup is a completion for points. >> >> In an election each matchup is also a competition for points ... in the >> form of ........ (you guessed it) votes! >> >> The scores we've been talking about are the votes in the head-to-head >> matchups. >> >> The beauty of RCV ballots is that for each matchup you can figure out the >> scores for both candidates from the RCV ballots. >> >> Suppose the matchup in question is between candidates X and Y. >> Separate the ballots into three stacks. >> >> Pretend that all of the other candidates are out of the picture so that >> all votes are transferred to X and Y. How many votes would X get? That is >> X's score for this matchup. >> >> You don't have to actually deface the ballots by crossing out the >> irrelevant candidates to find X's vote totals. Just count the number of >> ballots on which candidate X outranks candidate Y. >> >> This can be done for each of X's matchups, giving a complete set of >> matchup scores for X. >> >> Similarly we can get a completes set of matchup scores for each of the >> other candidates. >> >> And the one whose worst score is worse than anybody elses's worst score >> is the Gross Loser. >> >> So, to recap the method ... in the very first step eliminate the Gross >> Loser. In the next step eliminate the Gross Loser from among the remaining >> candidates, those who were not eliminated in the first step. >> >> In all subsequent steps where more than two candidates remain, eliminate >> the Gross Loser from among those remaining candidates. >> >> When it gets down to two candidates eliminate the Gross Loser, the one >> with the lowest score, i.e. the one with the fewest votes, i.e. elect the >> one with the most votes in this final matchup. >> >> This spells it out as plainly as I can do in the abstract. But what we >> all know is children do not learn how to play a game by reading the >> instructions on the inside cover of the box. They learn by playing with >> friends who already know the rules. >> >> This little card game (ballot counting game) is much simpler than >> Monopoly, Poker, Uno, Clue, etc that kids feel pretty confident with after >> a couple of dry runs with their friends. >> >> A YouTube video is the second best way to teach it. >> >> An EM text message is the worst way to teach it ... but you guys pick >> things up faster than average! >> >> -Forest >> >> >> >> >> >> >> On Thu, Nov 10, 2022, 9:07 AM Toby Pereira <tdp201b@yahoo.co.uk> wrote: >> >>> So do you have a nice and simple definition of this method that anyone >>> can understand? >>> >>> Where do you now stand on your Quick and Clean Burial Resistant Smith >>> method? At the time, it seemed to be the best thing since sliced bread, but >>> amongst all the posts, it now it appears not to have resisted, er, burial. >>> >>> Toby >>> >>> On Wednesday, 9 November 2022 at 22:07:53 GMT, Forest Simmons < >>> forest.simmons21@gmail.com> wrote: >>> >>> >>> I forgot to mention that Gross Loser Elimination is just as burial >>> resistant and Chicken resistant as IRV, and is less susceptible to >>> compromise than IRV, because unlike IRV, it has no Central Squeeze >>> pathology. >>> >>> Imagine candidates X and Y close to the left and right of Center Z. >>> Under sincere ranked ballots Z will have few first choice votes compared to >>> X and Y, so it will be eliminated, unless one of the factions compromises >>> and votes its second choice Z over its favorite. >>> >>> Which one would benefit by that insincere order reversal? >>> >>> Answer: the pairwise loser in the final runoff step between X and Y. >>> >>> A note on counting GLE.... a rectangular table of pairwise counts is >>> projected on the screen in the public counting room. >>> >>> The k_th entry in the j_th row of the table is the number of ballots on >>> which the j_th candidate out ranks the k_th candidate. >>> >>> As the ballots are opened and the candidate rankings carefully compared >>> one-by-one, the respective table entries for row j are incremental for each >>> candidate k that candidate j outranks on that ballot. >>> >>> When the ballots have been fully tabulated, the elimination steps begin. >>> >>> At each step the smallest entry in the table is circled. All viewers >>> must agree that it is indeed the smallest entry before continuing the step. >>> >>> Once all observers are in agreement that the smallest entry is the k_th >>> entry of row j, then candidate j is declared to be the Gross Loser of this >>> step, and so is eliminated by crossing out both the j_th row and the j_th >>> column of the table. >>> >>> The remaining table has one fewer row and one fewer column. >>> >>> Find the Gross Loser of this smaller table by identifying which row has >>> the smallest entry, etc. >>> >>> The last candidate standing is the GLE winner. >>> >>> If you want the frosting on the cake, have a representative for each >>> candidate announce if they claim to have the highest uncovered candidate in >>> the finish order. >>> >>> Process these claims in the reverse order, beginning with the GLE >>> winner, then the runner up, etc until either a claim is verified, or all >>> have been checked and refuted. >>> >>> To check a claim X, those who challenge X must produce a candidate Y who >>> beats X, but is not at the end of a two step beat path from X to Y. >>> >>> If the challengers cannot successfully refute the claim in this manner, >>> then the claim stands approved, and X is the winner. >>> >>> In other words, elect the candidate with the first unrefutted claim in >>> the order of claim processing ... which (as we have already specified) is >>> the reverse of the elimination order. >>> >>> Anybody have a better suggestion? >>> >>> Nobody? >>> >>> OK, then...how do we get the proposal ball rolling? >>> >>> -Forest >>> >>> >>> On Wed, Nov 9, 2022, 8:50 AM Forest Simmons <forest.simmons21@gmail.com> >>> wrote: >>> >>> This same simple tweak works on any method with a built in finish order, >>> including any one-at-a-time elimination method like IRV, BTR-IRV, Baldwin, >>> etc: >>> >>> Elect the uncovered candidate highest in the finish order. >>> >>> Why does our suggested tweak say to elect the highest uncovered >>> candidate in the finish order, instead of the highest unbeaten candidate in >>> the finish order? >>> >>> Answer: because sometimes there is no unbeaten candidate, but there is >>> always an uncovered candidate. >>> >>> The simplest and best one-by-one elimination method is Gross Loser >>> Elimination. >>> >>> No other one-at-time elimination method can improve on it, much less the >>> uncovered version: >>> >>> Elect the uncovered candidate highest in the Gross Elimination finish >>> order. >>> >>> Like IRV it is clone free. Unlike IRV it is precinct summable on one >>> pass through the ballots at each precinct. >>> >>> Wouldn't that have been nice last night at the midterm election count? >>> >>> Like IRV it is non monotonic, but unlike IRV it is Yee/Bolson monotonic: >>> the win regions are convex, not pathological fractals. [I almost wrote >>> Bolsonaro instead of Bolson ... sorry Brian!] >>> >>> Pick any method X, and pair it with Gross Loser Elimination ... >>> uncovered version or not ... and do a pairwise runoff between the two >>> winners. >>> >>> Not only will Gross Loser Elimination almost always come out ahead, the >>> people who do the experiment will come away saying, "Why do we even bother >>> with method X? GLE is so much more simple and effective." >>> >>> GLE is already Smith efficient without the uncovered tweak ... that's >>> just optional frosting on the cake. >>> >>> It is the simplest Smith efficient method that does not require >>> computing pairwise wins or losses. No need to mention Smith or Condorcet or >>> pairwise defeats. >>> >>> It automatically eliminates the Condorcet Loser at any stage when there >>> is one, because when there is a Condorcet Loser, it will also be the Gross >>> Loser. >>> >>> The Gross Loser is the candidate with the fewest ballots preferring it >>> over any other candidate. In a tournament, it is the candidate with the >>> single most embarrassingly low score. >>> >>> In fact, unlike IRV, Gross Loser Elimination can be used to get a finish >>> order for a Round Robin Tournament, so the uncovered tweak can be applied >>> to it if so desired. >>> >>> Suppose when there are only three uneliminated teams, team Rock's scores >>> against the other two teams stand at 60 and 40, while team Paper's scores >>> are 45 points against one team, and 72 against the other, and finally team >>> Scissors' scores stand at 35 and 90. >>> >>> Which team will be eliminated at this stage of GLE? >>> >>> Answer ... Scissors, because no other team scored as low as 35. >>> >>> Note that we did not even need to know who the other team was that >>> skunked Scissors, or how much it scored in that game to know that Scissors >>> was the Gross Loser of that round. >>> >>> Now tell me, who was the IRV loser of that round? >>> >>> Answer: impossible to know, because IRV makes no sense in a tournament >>> context, unless it is a superficial popularity contest of some kind. >>> >>> Is this the best RCV public proposal? >>> >>> No other Universal Domain method this simple is anywhere near as good. >>> >>> How about outside the UD? Do you think STAR is a better proposal? If so >>> why? >>> >>> -Forest >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> On Wed, Nov 9, 2022, 12:05 AM Forest Simmons <forest.simmons21@gmail.com> >>> wrote: >>> >>> In this context the most relevant question is what do we mean by >>> "uncovered", since that's the word used in the method definition ... >>> >>> Repeatedly eliminate the (remaining) candidate with fewest votes until >>> there remains only one uncovered candidate to elect. >>> >>> No need to know what covering means, although you can figure it out >>> indirectly from the definition of "uncovered:" >>> >>> A candidate is uncovered iff it has a beatpath of only two steps to each >>> candidate (if any) that beats it. >>> >>> Any candidate X who complains that they should have won because they >>> beat the winner W pairwise will get this truthful and obviously relevant >>> rejoinder: >>> >>> When you were eliminated, you had fewer transferred votes than I. >>> >>> I fact, I beat every candidate pairwise that was not already eliminated >>> (like you) on the basis of two few (transferred) votes. >>> >>> It is very easy to discern if some candidate X is uncovered: >>> >>> Just check each candidate Y that beats it (X) to see if it has a two >>> step beatpath via some Z, back to Y: >>> >>> X beats Z beats Y >>> >>> Only Smith candidates can be uncovered because only Smith candidates >>> have beatpaths back to the candidates that beat them. So the candidates you >>> have to check are the Smith candidates ... at most three, and rarely more >>> than one, in a public election. >>> >>> If you want, you can run IRV all the way through ... then if the IRV >>> winner is uncovered, you are done. If not, back up until you cone to an >>> uncovered candidate ... that's your winner! >>> >>> It's just a matter of doing regular IRV, and backing up (if necessary) >>> until you get to an uncovered candidate. >>> >>> Forest >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> On Tue, Nov 8, 2022, 11:18 AM Kristofer Munsterhjelm < >>> km_elmet@t-online.de> wrote: >>> >>> On 08.11.2022 18:02, Richard, the VoteFair guy wrote: >>> > Forest, what do you mean by "covered"? Is there a Wikipedia or >>> > Electowiki article (or section of an article) that explains it? Or is >>> > there a dictionary reference you can point to? >>> > >>> > Yes, you've used the words "covered" and "uncovered" many times but I >>> > don't recall ever seeing a clear explanation of what you mean. I >>> > presume it involves pairwise counts, but that's as far as I can guess. >>> >>> The short answer is: A covers B if A pairwise beats everybody B pairwise >>> beats and then some. >>> >>> An uncovered candidate is someone who is not covered by anyone else. >>> >>> This definition works when there are no pairwise ties. Things get >>> trickier with pairwise ties, as I found out when generalizing Friendly >>> Cover. >>> >>> -km >>> ---- >>> Election-Methods mailing list - see https://electorama.com/em for list >>> info >>> >>> ---- >>> Election-Methods mailing list - see https://electorama.com/em for list >>> info >>> >>

It may surprise you that given the same agenda, SPE (Sequential Pairwise Elimination) and Highest Uncovered are extremely hard to differentiate. There has probably never been an actual ballot set where they would elect a different candidate from the same agenda order. Certainly none of Kevin's seven taxonomical examples could do it because they all involve Smith sets of only three members ... in which case it is easy to show that the two methods will always agree. Here 's the demonstration: First, both methods are ISDA. SPE is IPDA because nonSmith candidates have no influence on the pairwise sorting of Smith, which always comes out solid ahead of the other candidates no matter what the sorting priority rule ... top, bottom, margins, etc. Highest Uncovered is ISDA because only Smith members can be uncovered ... so electing the highest uncovered agenda item is the same as electing the highest uncovered Smith item on the agenda. This brings us the next point: when there are only three Smith members, all of them are uncovered. This is because all Smith members have beatpaths to all other Smith members, and when there are few Smith members those beatpaths must be short (or unnecessarily repetitious)... in this case only one or two steps are possible without unnecessary repetition. The only remaining questionis which member of a three member Smith does SPE elect. Suppose the three members are X , Y and Z, and that their cyclic order is XYZX. If the agenda order is in agreement with cyclic order, pairwise sorting will not change it, so the highest agenda Smith candidate is the SPE winner. What if the agenda order is not the pairwise cyclic order? Then, since SPE goes from bottom to top, it will switch the order of the bottom two Smith members. That switch puts Smith into cyclic order so no further change in the order of Smith. The highest agenda Smith candidate gets elected, as in Highest Uncovered. In sum, when Smith has three members, SPE and Highest Uncovered both elect the top Smith candidate in the sgenda. Since SPE and Highest Uncovered are both equally easy to compute, the main question is what is the best agenda? Colin proposed basing the agenda on first place votes, but that is subject to note splitting (our only good reason for abandoning simple, traditional FPP Plurality) Here's an example that shows the vote splitting problem: 21 A1>B>C 19 A2>B>C 35 B>C>A1 33 C>A2>A1 The agenda order is B>C>A1>A2. B is both the SPE winner and the Highest Uncovered agenda winner, as well as the winner of Ranked Pairs, River, Schulze, and many other methods including IRV. If A had not been split into A1 and A2, the agenda would have been A>B>C, and the winner of all the aforementioned methods would have been A. So that's why I suggested Implici Approval: If they are not going to allow us to rank more than three candidates, then the simplest agenda method that avoids vote splitting is the implicit approval agenda ... the candidates most favored by the agenda are the ones unranked on the fewest ballots. Since you cannot rank all of the candidates leave the candidates that least approve of unranked. Although that's the simplest acceptable agenda in this context (of forced truncations) there may be another that is worth the extra complexity. -Forest On Sat, Nov 12, 2022, 12:00 AM Forest Simmons <forest.simmons21@gmail.com> wrote: > Evidently the IRV proposers hacve had to settle for ránking only there or > tour candidateson each ballot ... better than nothing. > > One pass through the ballots to get the pairwise information and the > number of truncations for each candidate ... the exact same work as the SPE > method you propose ... but an agenda of first place votes breeds lots of > vote splitting unless you expect the voters to have lots of equal first > rankings ... not a good idea. > > Electing the uncovered candidate unranked on the fewest ballots is a > simpler Condorcet method than SPE ... and it is guaranteed to elect an > uncovered member of the Smith Set. > > But SPE is also good ... if the agenda is clone independent, like implicit > approval. > > -Forest > > > On Fri, Nov 11, 2022, 5:26 PM Colin Champion < > colin.champion@routemaster.app> wrote: > >> On 11/11/2022 20:49, Forest Simmons wrote: >> > >> > Since almost all RCV implementations limit the number of candidates >> > that can be ranked on a ballot, the simplest decent RCV method is ... >> > Elect the uncovered candidate that is unranked on the fewest ballot >> > >> Does anyone know why this truncation is imposed? If it’s to limit the >> amount of work needed to count the ballots, wouldn’t it make sense for >> Condorcet supporters to advocate a method which was countable in linear >> time? In practice this would presumably be Sequential Pairwise >> Elimination with an FPTP pre-ranking. If you insist on a quadratic time >> method and accept the corollary of ballot truncation, I don’t imagine it >> will work very well. Or am I missing something? >> >> CJC >> ---- >> Election-Methods mailing list - see https://electorama.com/em for list >> info >> >

On 14/11/2022 03:54, Forest Simmons wrote.. > many wise things I think Forest slightly missed my point - I was arguing that SPE could be proposed *as an alternative* to truncation, since I surmised (incorrectly) that truncation might have been justified by a desire to avoid quadratic counting costs. Hence there would be no cutoff as part of the voting procedure, and no natural concept of implicit approval. If there is political opposition to mechanical counting, then proposing a linear-time countable method seems to me better than either putting up with truncation or fighting unnecessary battles. I don't know why linear-time countability dropped out of consideration. In 1275 Llull proposed the "league table" method now often named after Copeland, but with one difference - he didn't suggest ranked preference voting (after all, paper hadn't been invented then, let alone computers). Instead he called for m(m-1)/2 pairwise subelections. In 1299, presumably recognising the unworkability of this, he proposed a "knockout" method comprising just m-1 pairwise subelections which amounts to SPE with an arbitrary agenda. I think this was in one of his writings which got lost.    A little later, taking advantage of the invention of paper, Nicholas of Cusa proposed the Borda count in conjunction with ranked preference voting. Borda reinvented his ideas in the 18th century. Condorcet (in his Essai) got bogged down in problems arising from cycles, and partially extricated himself with a quadratic-time algorithm which is a simplified ranked pairs. In his later writings he reinvented Llull's league table method and instantly rejected it on account of its quadratic cost. In its place he advocated what I take to be a defective form of Bucklin's method.[1] This was adopted in Geneva and "found not to work" (I don't know the details, but Bucklin's method is quite a large step backwards).    Nanson carried on from Condorcet. He agreed that the quadratic cost made Llull's league table unworkable, wasn't attracted to Bucklin's method, and instead proposed his own. He may have been a little optimistic in his costings, but at worst his method is m log m.    That seems to be the end of the discussion. When Black proposed his method in the middle of the last century, voting technology was no different than in Condorcet's day, but he doesn't seem to have worried about the counting costs. Llull's knockout reappeared as SPE with a pre-ranking stage to eliminate the obvious asymmetry, but since quadratic-time pre-rankings are often assumed, counting time hasn't always been the main consideration.    CJC [1]. I always confuse Bucklin's method with Baldwin's. I wish the latter was called the "Trinity College" method, since it was invented before Baldwin's time. The Trinity College Dialectic Society was founded at Nanson's university a few years before Nanson wrote his memoir, so its voting method was presumably an early version of Nanson's.