> On 03/01/2024 10:49 PM EST Michael Ossipoff <email9648742@gmail.com> wrote: > ... > > The entire Participation-check only takes half as long as the original exhaustive pairwise-count used by Condorcet. > I have to admit that I am not following this closely, but in my superficial understanding of the argument, we're discussing the complexity of the tabulation as if it were done centrally, by a single computer, similarly to what we **have** to do with Hare RCV. But, essentially, every Condorcet-consistent method is Precinct Summable with N(N-1)/2 pairs of numbers (N is the number of candidates) that can be computed locally at each polling place and reported upward to be added up, similarly to FPTP (except the latter needs only N numbers). In fact those N(N-1) tallies can be incremented as each ballot is inserted into the voting tabulator at the precinct level. If first-choice votes are also tallied (say, if the method enacted is Condorcet-Plurality), the number of tallies increases from N(N-1) to simply N². But the hard work is done decentralized by *many* computers. It's distributed computation. FPTP, Approval, Score all require fewer tallies than does Condorcet RCV. STAR is also N². But Hare RCV is floor((e-1)N!)-1 which is much worse, essentially proscribing local precinct tabulation and yet there are statewide RCV elections (that take two weeks to report the election outcome). I just don't see what the problem is other than some theoretical academic navel gazing. -- r b-j . _ . _ . _ . _ rbj@audioimagination.com "Imagination is more important than knowledge." . . .

Robert: I already said that the negligibly tiny possibility of one a particular voter’s voted-preference being negatively responded-to doesn’t even begin to compare in importance—from that voter’s point of view—to the complete strategy-freedom of RP(wv). But we were speaking of how Condorcet could be used in Germany, in compliance with Germany’s Constitution. Checking for, detecting & discarding a Participation-violating outcome, & then counting the ballots by Implicit-Approval, which never responds negatively, was the suggestion being discussed. On Fri, Mar 1, 2024 at 22:06 robert bristow-johnson < rbj@audioimagination.com> wrote: > > > > > > > The entire Participation-check only takes half as long as the original > exhaustive pairwise-count used by Condorcet. > > > > I have to admit that I am not following this closely, but in my > superficial understanding of the argument, we're discussing the complexity > of the tabulation as if it were done centrally, by a single computer, > similarly to what we **have** to do with Hare RCV. Then your superficial understanding of the argument is incorrect. I said nothing about where the individual pairwise votes are counted. I said that a Participation-violation-check requires the counting of only half as many individual pairwise votes, & only takes half as long, compared to the election’s original exhaustive pairwise-count. …the veracity of which is unaffected by where the counting is done. …unless you want to count one centrally, & the other at precincts. :-) I gave numbers for how many individual pairwise votes need to be counted, in the original exhaustive pairwise count, & in the complete thorough Participation-violation check, with 300 million voters & 20 candidates. …which is unaffected by where the counting is done > > > > > FPTP, Approval, Score all require fewer tallies than does Condorcet RCV. > STAR is also N². But Hare RCV is floor((e-1)N!)-1 which is much worse, > essentially proscribing local precinct tabulation and yet there are > statewide RCV elections (that take two weeks to report the election > outcome). > Though of course the number of votes per votes needing to be counted, for a given number of candidates, varies in Approval, & varies in Hare, it seemed to me, when I looked at it, that it’s roughly about the same for the 2 methods. That first impression of mine could have been mistaken. But, if so, it’s one of the reasons why Hare, with a remarkably easy handcount for a rank-method, would be a good choice for an informal vote on a nonpolitical matter like a pizza topping or movie choice. > > I just don't see what the problem is other than some theoretical academic > navel gazing. …if a Constitutional-prohibition is some theoretical academic naval gazing. > > For someone who doesn’t see what the problem is, you seem bent on making one. > -- > > r b-j . _ . _ . _ . _ rbj@audioimagination.com > > "Imagination is more important than knowledge." > > . > . > . > ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

Damn; replacing pairwise majorities with pairwise *Golden Ratios *recovers a property slightly weaker than full participation: https://link.springer.com/article/10.1007/s10726-014-9416-4 But more relevant is that quota-Condorcet satisfies participation (for all voter profiles) iff we set quota = (n_candidates - 1) / n_candidates: https://www.sciencedirect.com/science/article/pii/0166218X88900881

On Fri, Mar 1, 2024 at 21:48 Closed Limelike Curves < closed.limelike.curves@gmail.com> wrote > Agree that if RP improves strategy-resistance in practice, a tiny risk of > participation failures is worth it. But courts might not agree. > It’s a given that governments often do undesirable things. > More importantly, it means you can't challenge IRV in court on the basis > that participation/monotonicity failure violates a fundamental human right > (if your own proposed alternative also violates that right) > I don’t object to Hare’s negative-response. I prefer Condorcet to it, for public political elections because of RP(wv)’s freedom from strategy-need, & Hare’s potential for lesser-evil giveaway perceived strategy-need maybe for some people. If Hare has been enacted by a progressive electorate, then it’s a safe bet that they didn’t do so because they want to bury their favorite under a lesser evil… but rather because they DON’T want or intend to. So, Hare’s potential problem likely won’t manifest. …especially if it’s honestly sold…which of course it isn’t. What I object to about Hare is FairVote’s fraudulent promotion of it as something it isn’t. But I *do* criticize STV’s negative response, because Saint > > That occurred to me too. Well I’d say that the important thing is that >> after the illegal Participation-failure result is discarded, then even if >> IA gives the same winner, this time it didn’t happen as a >> Participation-violation. >> > It's still a participation violation--the whole combined electoral system > violates participation, even if the specific steps don't. In the same way, > there's no individual step of IRV that violates participation (each round > is just a plurality vote). It's the process as a whole that does. > > On Fri, Mar 1, 2024 at 7:49 PM Michael Ossipoff <email9648742@gmail.com> > wrote: > >> Yes, just a matter of, for each voter, decrementing the population-total >> of every pairwise-preference s/he voted. …& applying RP(wv) to that >> modified population preference-totals set. But, other than looking at that >> voter’s ballot, no new votecounting is needed. >> >> The entire Participation-check only takes half as long as the original >> exhaustive pairwise-count used by Condorcet. >> >> >> >>> Couldn't this new procedure still violate participation? >>> >> >> Not if it’s Implicit-Approval (ranked = approved). >> >> >> What if adding ballots[i] with A>B would cause B>A, but B also happens >>> to win under implicit approval voting? >>> >> >> That occurred to me too. Well I’d say that the important thing is that >> after the illegal Participation-failure result is discarded, then even if >> IA gives the same winner, this time it didn’t happen as a >> Participation-violation. >> >> The discarding makes the election innocent. >> >> But Steve is right: For any one of us, the minuscule chance of a >> preference of ours being negatively responded-to doesn’t even come close to >> outweighing the strategy-freeness for which RP(wv) is being used. >> >>> >>> >>> On Thu, Feb 29, 2024 at 4:02 PM Michael Ossipoff <email9648742@gmail.com> >>> wrote: >>> >>>> Yes, I too feel that though Consistency-failure is an embarrassment, >>>> it’s Participation-failure—where a particular voter should have stayed >>>> home—that is the rights-violation. >>>> >>>> Yes, maybe there’s some big subset of the voters, say 37%, such that, >>>> if they’d stayed home the result of applying the method to the other 63% >>>> (which would then have been the actual election-result) would have been the >>>> same as the result of applying it to them separately…but, because they >>>> voted, a different result happened. >>>> >>>> That doesn’t sound as bad as Participation-failure. >>>> >>>> 1. That 37%’s hypothetical collective-choice by the method isn’t as >>>> definite & stark as a person’s expressed-preference. >>>> >>>> 2. A *different outcome* from the hypothetical result of applying the >>>> method to the 37% isn’t as concretely clearly wrong as outright >>>> negative-response to someone’s expressed preference. >>>> >>>> …& that’s a good thing, because checking for Consistency-failure is >>>> what DOES sound computationally infeasible due to the many ways of dividing >>>> a large electorate into 2 par. >>>> >>>> Checking for Participation-failure would just be a matter of, for each >>>> voter, doing the count without hir ballot & determining whether that >>>> changes the winner to someone s/he ranked above the actual winner. >>>> >>>> With 300,000,000 voters that isn’t infeasible with today’s computers is >>>> it? >>>> >>>> I should have asked: >>>> >>>> In Germany, is Consistency-failure unconstitutional? That might be a >>>> problem for Condorcet. If only Participation-failure is unconstitutional, >>>> then wouldn’t Kobe computationally-feasible to check for that? >>>> >>>> When a Participation-failure is found, then elect the winner by >>>> Implicit-Approval. >>>> >>>> In wv, not ranking candidates you don’t like is good defensive-strategy >>>> anyway. If it’s known that many people vote that way, wv’s already good >>>> burial-deterrence is further enhanced. >>>> >>>> >>>> >>>> On Thu, Feb 29, 2024 at 10:06 Closed Limelike Curves < >>>> closed.limelike.curves@gmail.com> wrote: >>>> >>>>> That seems to exclude participation failures except in the case >>>>> of near-ties; I think adding a batch of ballots, all of which rank A>B, >>>>> shouldn't cause B to defeat A. >>>>> >>>>> I don't think we need perfect consistency, just participation. >>>>> Consistency failures are paradoxical, but consistency is a very >>>>> strong criterion and isn't needed for the weaker case of participation >>>>> (which limits consistency to the ). It's also strong enough that it >>>>> could lock you into Kemeny-Young (the unique order-consistent Condorcet >>>>> method). >>>>> >>>>> Checking for participation naively definitely isn't feasible, so we'd >>>>> need a more elegant approach. >>>>> >>>>> Alternatively, we could try finding a Condorcet system that violates >>>>> participation "as little as possible," in the sense that any participation >>>>> failures are forced by Condorcet-compliance. A judge might be willing >>>>> to sign off on that, since it's a tradeoff between two similarly-important >>>>> values (equal protection and majority rule). >>>>> >>>>> On Wed, Feb 28, 2024 at 10:30 PM Michael Ossipoff < >>>>> email9648742@gmail.com> wrote: >>>>> >>>>>> Maybe it might be good enough to just require that there must not be >>>>>> a Participation-violation with respect to any one of the voters, had s/he >>>>>> voted last? >>>>>> >>>>>> On Wed, Feb 28, 2024 at 22:17 Michael Ossipoff < >>>>>> email9648742@gmail.com> wrote: >>>>>> >>>>>>> Well I don’t know if the Consistency-check would be >>>>>>> Computationally-feasible, because of course there are a lot of ways to >>>>>>> divide a large electorate into 2 parts. >>>>>>> >>>>>>> That might be one more good reason to use Approval instead, for >>>>>>> those single-winner elections. >>>>>>> >>>>>>> On Wed, Feb 28, 2024 at 22:07 Michael Ossipoff < >>>>>>> email9648742@gmail.com> wrote: >>>>>>> >>>>>>>> I call RP(wv) with that modification “Nonsense-Free RP(wv)”, & >>>>>>>> propose it for Germany’s single-winner elections, including the single >>>>>>>> member district elections in their Additional-Member proportional >>>>>>>> topping-up Parliamentary elections. >>>>>>>> >>>>>>>> On Wed, Feb 28, 2024 at 21:47 Michael Ossipoff < >>>>>>>> email9648742@gmail.com> wrote: >>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> On Wed, Feb 28, 2024 at 21:07 Closed Limelike Curves < >>>>>>>>> closed.limelike.curves@gmail.com> wrote: >>>>>>>>> >>>>>>>>>> . Some alternative criterion that gets us "99% of the way to >>>>>>>>>>> Condorcet," so it behaves like Condorcet except in the rare cases where it >>>>>>>>>>> conflicts with participation (or maybe just mono-add-top/remove-bottom). >>>>>>>>>> >>>>>>>>>> >>>>>>>>> There might be better ways, but there’s always the lexicographical >>>>>>>>> way. The criterion could require that Participation (& other >>>>>>>>> non-opposite-response criteria) & Consistency be met. …& that the voted >>>>>>>>> CW, when there is one must be elected when that doesn’t conflict with the >>>>>>>>> above requirements. >>>>>>>>> >>>>>>>>> A complying method could just repeat that wording, along with a >>>>>>>>> specification about what to do if there’s no voted CW, & what to do if the >>>>>>>>> ballot-configuration is such that additional of a new ballot could violate >>>>>>>>> Participation, or if some division of the electorate into 2 parts could >>>>>>>>> show a Consistency violation. >>>>>>>>> >>>>>>>>> Maybe apply Implicit-Approval to the ballots then. (Ranked = >>>>>>>>> approved) >>>>>>>>> >>>>>>>>>> >>>>>>>>>> On Wed, Feb 28, 2024 at 8:57 PM Closed Limelike Curves < >>>>>>>>>> closed.limelike.curves@gmail.com> wrote: >>>>>>>>>> >>>>>>>>>>> It’s surprising that participation-violation is unconstitutional >>>>>>>>>>>> in Germany, because, here, even Hare’s greater nonmonotonicity is okay. >>>>>>>>>>>> >>>>>>>>>>> I'm actually not sure it is--the Supreme Court has never ruled >>>>>>>>>>> on , and courts also haven't ruled on the constitutionality of non-monotone >>>>>>>>>>> voting rules. STV has been upheld as constitutional in the past, but the >>>>>>>>>>> challenges were never brought over monotonicity failures. It's entirely >>>>>>>>>>> possible a new challenge could overturn it; there's a strong argument that >>>>>>>>>>> monotonicity failures violate due process and the equal protection clause. >>>>>>>>>>> >>>>>>>>>>> The ideal case to bring to the Supreme Court would have been for >>>>>>>>>>> Begich's campaign to sue after the 2022 Alaska election. A moderate >>>>>>>>>>> Republican plaintiff is appealing to the mostly-Republican Supreme >>>>>>>>>>> Court, without being too controversial. Being the Condorcet winner makes >>>>>>>>>>> his case look even stronger. >>>>>>>>>>> >>>>>>>>>>> On the other hand, if someone says the word "monotonicity" in >>>>>>>>>>> front of a judge, their eyes will glaze over and they'll immediately stop >>>>>>>>>>> caring about all this weird, complicated nerd math. The way to explain >>>>>>>>>>> participation failures is to run a ton of ads explaining to Alaska >>>>>>>>>>> Republicans that Begich lost because *he got* *too many votes. * >>>>>>>>>>> >>>>>>>>>>> One suggestion: why not rename monotonicity to "helpfulness?" >>>>>>>>>>> (Voting should help your candidate, not hurt them). We can call >>>>>>>>>>> monotonicity failures "spitefulness" (because the system is going out of >>>>>>>>>>> its way to do the opposite of what you ask it to). >>>>>>>>>>> >>>>>>>>>>> On Wed, Feb 28, 2024 at 11:32 AM Michael Ossipoff < >>>>>>>>>>> email9648742@gmail.com> wrote: >>>>>>>>>>> >>>>>>>>>>>> It’s surprising that participation-violation is >>>>>>>>>>>> unconstitutional in Germany, because, here, even Hare’s greater >>>>>>>>>>>> nonmonotonicity is okay. >>>>>>>>>>>> >>>>>>>>>>>> It’s disingenuous to say that Hare is nonmonotonic & Condorcet >>>>>>>>>>>> isn’t. Nonmonotonicity is just defined to give Condorcet, with it’s >>>>>>>>>>>> participation-failure, a pass. >>>>>>>>>>>> >>>>>>>>>>>> I’ve heard that Participation & the Condorcet Criterion are >>>>>>>>>>>> mutually incompatible. I feel that participation-failure is an acceptable >>>>>>>>>>>> price for the Condorcet Criterion. Always electing the voted CW brings >>>>>>>>>>>> strategy improvement, & the unpredictable & rare participation-failure is >>>>>>>>>>>> probably irrelevant to strategy. >>>>>>>>>>>> >>>>>>>>>>>> But that incompatibility, along with the ones Arrow >>>>>>>>>>>> pointed-out, shows that single-winner elections aren’t perfect. …making a >>>>>>>>>>>> good argument for PR…*monotonic* PR, which excludes STV & Largest-Remainder. >>>>>>>>>>>> >>>>>>>>>>>> Maybe, as a PR country (like 2/3 of the world’s countries), >>>>>>>>>>>> Germany feels no need to compromise participation. >>>>>>>>>>>> >>>>>>>>>>>> We’re told that list-PR “hasn’t been tried”. No, just in 2/3 of >>>>>>>>>>>> the world’s countries for about a century. >>>>>>>>>>>> >>>>>>>>>>>> But, with that counterfactual “hasn’t been tried” excuse, we’re >>>>>>>>>>>> stuck in the 18th century, & always will be, while most of the world has >>>>>>>>>>>> moved on to democracy. >>>>>>>>>>>> >>>>>>>>>>>> On Wed, Feb 28, 2024 at 10:36 Closed Limelike Curves < >>>>>>>>>>>> closed.limelike.curves@gmail.com> wrote: >>>>>>>>>>>> >>>>>>>>>>>>> Can Condorcet be weakened to comply with participation? >>>>>>>>>>>>> Condorcet methods have plenty of advantages, but systems failing >>>>>>>>>>>>> participation are vulnerable to court challenges or being struck down as >>>>>>>>>>>>> unconstitutional, as seen in Germany. >>>>>>>>>>>>> ---- >>>>>>>>>>>>> Election-Methods mailing list - see https://electorama.com/em >>>>>>>>>>>>> for list info >>>>>>>>>>>>> >>>>>>>>>>>>

Hi Mike and everyone, First off, if anyone was missing my site, it is back up. I had to find different hosting (a bit abruptly). Where I was trying to link right before the site went down: votingmethods.net/cond works out a given (or random) scenario for Schulze, RP, or River. You just have to expand sections at the bottom of the result. So it could be worth a look. Mike wrote: > Is River as easy to define &. explain as RP?. I see I should try to write out clearly how I suggest to understand River. There is no "final ranking" in River. Instead every candidate begins "below no one" or "subordinated to no one." This is sort of a ranking but the "trees" we make go only one level down: you will never be able to ascend two positions from a given candidate. 1. Initially each candidate is subordinated to no one. 2. Consider each pairwise defeat from strongest to weakest. 3. When you consider a defeat, ask whether the loser is subordinated to anyone? If so: Ignore the defeat and proceed to the next. If not, then ask: 4. Is the defeat winner subordinated to the defeat loser? If so, ignore the defeat and go to the next. 5. Is the defeat winner subordinated to someone else? If so, the defeat loser, along with everyone subordinated to them, becomes subordinated to the candidate that the defeat winner is subordinated to. 6. Otherwise it must be that the defeat winner is subordinated to no one. So here the defeat loser, along with everyone subordinated to them, becomes subordinated to the defeat winner. 7. End loop. Go to the next defeat. 8. In the end, the candidates subordinated to no one are the winners. Alternatively instead of talking about subordination, you can say that each candidate has their own "bin" and starts in their own and may move to another. This would allow you to merge steps 4 and 5: "4. The defeat loser, along with everyone *in the loser's bin*, moves to whichever bin the defeat winner is currently located in." And if the latter bin happens to be the loser's bin, in effect nothing happens. We don't need a rule saying to ignore the defeat, because the bin movement doesn't change anything either. I can understand if a reader eyeballs all that and says this looks like a mess and it's not clearer than RP. But hear me out on the *ease* of it: 1. If you are programming River, you never actually check for a cycle, whether a proposed defeat would create one. And comparing to Schulze, you never trace a beatpath or find its strength, or (by its other algorithm) have to find the Schwartz set repeatedly. 2. If you are solving it by hand, it would be enough to have a fridge magnet for each candidate, start them out in imaginary bins, and push the magnets around in a straightforward way to track who is subordinated to whom. It may be possible to define RP more concisely, but it takes some work to figure out what it is actually saying to do to solve it. Hopefully the above explains it better than I have before. Kevin votingmethods.net

Wow. River doesn’t need the exhaustive pairwise-count? How does its count-time compare to that of RP? I didn’t know that about River. I believed that only Sequential-Pairwise was the only exception to the need for the exhaustive pairwise-count. The exhaustive count requires, per voter, counting one pairwise-vote for each possible pair of candidates. How many votes need to be counted per voter in River? If one only cares about finding the winner, rather than an output-ranking, could the count-instruction be written more briefly? As written, it’s much too complicated for a public-proposal. Someone said that River is better at deterring burial. I disagree. It seems to me that skipping a defeat if its defeated is defeated in an already-kept defeat undermines autodeterence. Only one of the CW’s defeats is kept. That means that every Bus but one can’t have its defeat dropped, so only one Bus survives. I like it if the exhaustive pairwise-count isn’t needed, but can the count-instructions be written more briefly, if only the winner is needed? On Sat, Mar 9, 2024 at 07:02 Kevin Venzke <stepjak@yahoo.fr> wrote: > Hi Mike and everyone, > > First off, if anyone was missing my site, it is back up. I had to find > different > hosting (a bit abruptly). > > Where I was trying to link right before the site went down: > votingmethods.net/cond > works out a given (or random) scenario for Schulze, RP, or River. You just > have to > expand sections at the bottom of the result. So it could be worth a look. > > Mike wrote: > > Is River as easy to define &. explain as RP?. > > I see I should try to write out clearly how I suggest to understand River. > > There is no "final ranking" in River. Instead every candidate begins > "below no one" > or "subordinated to no one." This is sort of a ranking but the "trees" we > make go > only one level down: you will never be able to ascend two positions from a > given > candidate. > > 1. Initially each candidate is subordinated to no one. > 2. Consider each pairwise defeat from strongest to weakest. > 3. When you consider a defeat, ask whether the loser is subordinated to > anyone? > If so: Ignore the defeat and proceed to the next. > If not, then ask: > 4. Is the defeat winner subordinated to the defeat loser? If so, ignore > the defeat > and go to the next. > 5. Is the defeat winner subordinated to someone else? If so, the defeat > loser, along > with everyone subordinated to them, becomes subordinated to the candidate > that the > defeat winner is subordinated to. > 6. Otherwise it must be that the defeat winner is subordinated to no one. > So here > the defeat loser, along with everyone subordinated to them, becomes > subordinated to > the defeat winner. > 7. End loop. Go to the next defeat. > 8. In the end, the candidates subordinated to no one are the winners. > > Alternatively instead of talking about subordination, you can say that each > candidate has their own "bin" and starts in their own and may move to > another. > This would allow you to merge steps 4 and 5: > "4. The defeat loser, along with everyone *in the loser's bin*, moves to > whichever > bin the defeat winner is currently located in." > And if the latter bin happens to be the loser's bin, in effect nothing > happens. We > don't need a rule saying to ignore the defeat, because the bin movement > doesn't > change anything either. > > I can understand if a reader eyeballs all that and says this looks like a > mess and > it's not clearer than RP. > > But hear me out on the *ease* of it: > > 1. If you are programming River, you never actually check for a cycle, > whether a > proposed defeat would create one. And comparing to Schulze, you never > trace a > beatpath or find its strength, or (by its other algorithm) have to find > the Schwartz > set repeatedly. > 2. If you are solving it by hand, it would be enough to have a fridge > magnet for > each candidate, start them out in imaginary bins, and push the magnets > around in a > straightforward way to track who is subordinated to whom. > > It may be possible to define RP more concisely, but it takes some work to > figure out > what it is actually saying to do to solve it. > > Hopefully the above explains it better than I have before. > > Kevin > votingmethods.net >

River is actually really easy to explain: 1. List all pairwise matches from biggest to smallest margin of victory. 2. If a candidate loses a match, cross them out (declare them to be a loser). Cross out any redundant matches that involve them (anything that would make them get eliminated twice). 3. Cross out any elections that would create a cycle. On Sat, Mar 9, 2024 at 2:46 PM Michael Ossipoff <email9648742@gmail.com> wrote: > Wow. River doesn’t need the exhaustive pairwise-count? How does its > count-time compare to that of RP? > > I didn’t know that about River. I believed that only Sequential-Pairwise > was the only exception to the need for the exhaustive pairwise-count. > > The exhaustive count requires, per voter, counting one pairwise-vote for > each possible pair of candidates. > > How many votes need to be counted per voter in River? > > If one only cares about finding the winner, rather than an output-ranking, > could the count-instruction be written more briefly? > > As written, it’s much too complicated for a public-proposal. > > Someone said that River is better at deterring burial. I disagree. It > seems to me that skipping a defeat if its defeated is defeated in an > already-kept defeat undermines autodeterence. > > Only one of the CW’s defeats is kept. That means that every Bus but one > can’t have its defeat dropped, so only one Bus survives. > > I like it if the exhaustive pairwise-count isn’t needed, but can the > count-instructions be written more briefly, if only the winner is needed? > > On Sat, Mar 9, 2024 at 07:02 Kevin Venzke <stepjak@yahoo.fr> wrote: > >> Hi Mike and everyone, >> >> First off, if anyone was missing my site, it is back up. I had to find >> different >> hosting (a bit abruptly). >> >> Where I was trying to link right before the site went down: >> votingmethods.net/cond >> works out a given (or random) scenario for Schulze, RP, or River. You >> just have to >> expand sections at the bottom of the result. So it could be worth a look. >> >> Mike wrote: >> > Is River as easy to define &. explain as RP?. >> >> I see I should try to write out clearly how I suggest to understand River. >> >> There is no "final ranking" in River. Instead every candidate begins >> "below no one" >> or "subordinated to no one." This is sort of a ranking but the "trees" we >> make go >> only one level down: you will never be able to ascend two positions from >> a given >> candidate. >> >> 1. Initially each candidate is subordinated to no one. >> 2. Consider each pairwise defeat from strongest to weakest. >> 3. When you consider a defeat, ask whether the loser is subordinated to >> anyone? >> If so: Ignore the defeat and proceed to the next. >> If not, then ask: >> 4. Is the defeat winner subordinated to the defeat loser? If so, ignore >> the defeat >> and go to the next. >> 5. Is the defeat winner subordinated to someone else? If so, the defeat >> loser, along >> with everyone subordinated to them, becomes subordinated to the candidate >> that the >> defeat winner is subordinated to. >> 6. Otherwise it must be that the defeat winner is subordinated to no one. >> So here >> the defeat loser, along with everyone subordinated to them, becomes >> subordinated to >> the defeat winner. >> 7. End loop. Go to the next defeat. >> 8. In the end, the candidates subordinated to no one are the winners. >> >> Alternatively instead of talking about subordination, you can say that >> each >> candidate has their own "bin" and starts in their own and may move to >> another. >> This would allow you to merge steps 4 and 5: >> "4. The defeat loser, along with everyone *in the loser's bin*, moves to >> whichever >> bin the defeat winner is currently located in." >> And if the latter bin happens to be the loser's bin, in effect nothing >> happens. We >> don't need a rule saying to ignore the defeat, because the bin movement >> doesn't >> change anything either. >> >> I can understand if a reader eyeballs all that and says this looks like a >> mess and >> it's not clearer than RP. >> >> But hear me out on the *ease* of it: >> >> 1. If you are programming River, you never actually check for a cycle, >> whether a >> proposed defeat would create one. And comparing to Schulze, you never >> trace a >> beatpath or find its strength, or (by its other algorithm) have to find >> the Schwartz >> set repeatedly. >> 2. If you are solving it by hand, it would be enough to have a fridge >> magnet for >> each candidate, start them out in imaginary bins, and push the magnets >> around in a >> straightforward way to track who is subordinated to whom. >> >> It may be possible to define RP more concisely, but it takes some work to >> figure out >> what it is actually saying to do to solve it. >> >> Hopefully the above explains it better than I have before. >> >> Kevin >> votingmethods.net >> > ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

Hi Closed Curves, Mike, I thought about the definition a little. Let's try this, using the "bins" concept rather than the "subordination" one: 1. Initially each candidate is in their own "bin" named for themselves. 2. Consider each pairwise defeat from strongest to weakest. 3. For a given defeat X>Y, move everyone in bin Y to the bin that X is in. 4. End loop. Go to the next defeat. 5. Elect one of the candidates who remains in their original bin. Things to note about this: Once Y has moved anywhere, no one will ever be in bin Y again. So we don't actually have to throw out other defeats over Y. If X is in Y's bin when X>Y is considered, rule 3 says that everyone in bin Y (including X) moves to where X is, which is already Y. So nothing will happen in this case either. Should be right. Closed Limelike Curves <closed.limelike.curves@gmail.com> a écrit : > River is actually really easy to explain: > 1. List all pairwise matches from biggest to smallest margin of victory. > 2. If a candidate loses a match, cross them out (declare them to be a loser). >Cross out any redundant matches that involve them (anything that would make them >get eliminated twice). > 3. Cross out any elections that would create a cycle. I'd note that River isn't usually defined using margin of victory. (Certainly, neither Mike nor I would do that.) But yes, normally River is explained the same way as RP except with an extra rule to ignore defeats over candidates already defeated. That might be more appealing. But what I want to point out is that if you are asking "would locking X>Y create a cycle?" for River, that is worsening the runtime and is harder to solve whether manually or programmatically. It is possible, I suppose, that one might use one algorithm to explain a proposal, or do the needed advocacy, but then use a second, different algorithm to solve it in practice. Michael Ossipoff <email9648742@gmail.com> wrote: > Wow. River doesn’t need the exhaustive pairwise-count? How does its count-time > compare to that of RP? >  > I didn’t know that about River. I believed that only Sequential-Pairwise was the > only exception to the need for the exhaustive pairwise-count. >  > The exhaustive count requires, per voter, counting one pairwise-vote for each > possible pair of candidates. > > How many votes need to be counted per voter in River? If I understand you correctly, River does need it. River still needs you to count all the preferences from each ballot. River's runtime (as I described River) should be proportional to the number of pairwise contests, so roughly N^2. For each contest, the action you take consumes a constant time. That is definitely faster than RP, because RP expends a lot of effort assessing potential cycles. > If one only cares about finding the winner, rather than an output-ranking, could > the count-instruction be written more briefly? Unfortunately, River as I described it only returns a winner already. Kevin votingmethods.net

On Sat, Mar 9, 2024 at 15:50 Closed Limelike Curves < closed.limelike.curves@gmail.com> wrote: > River is actually really easy to explain: > 1. List all pairwise matches from biggest to smallest margin of victory. > 2. If a candidate loses a match, cross them out (declare them to be a > loser). Cross out any redundant matches that involve them (anything that > would make them get eliminated twice). > 3. Cross out any elections that would create a cycle. > Could that be worded as: List the defeats, one at a time, stronger ones first. But skip any defeat that cycles with already-listed defeats. …& also skip any defeat whose defeated-candidate is defeated in an already-listed defeat. When all defeats have been listed or skipped, elect the candidate who isn’t beaten in a listed defeat. > > > On Sat, Mar 9, 2024 at 2:46 PM Michael Ossipoff <email9648742@gmail.com> > wrote: > >> Wow. River doesn’t need the exhaustive pairwise-count? How does its >> count-time compare to that of RP? >> >> I didn’t know that about River. I believed that only Sequential-Pairwise >> was the only exception to the need for the exhaustive pairwise-count. >> >> The exhaustive count requires, per voter, counting one pairwise-vote for >> each possible pair of candidates. >> >> How many votes need to be counted per voter in River? >> >> If one only cares about finding the winner, rather than an >> output-ranking, could the count-instruction be written more briefly? >> >> As written, it’s much too complicated for a public-proposal. >> >> Someone said that River is better at deterring burial. I disagree. It >> seems to me that skipping a defeat if its defeated is defeated in an >> already-kept defeat undermines autodeterence. >> >> Only one of the CW’s defeats is kept. That means that every Bus but one >> can’t have its defeat dropped, so only one Bus survives. >> >> I like it if the exhaustive pairwise-count isn’t needed, but can the >> count-instructions be written more briefly, if only the winner is needed? >> >> On Sat, Mar 9, 2024 at 07:02 Kevin Venzke <stepjak@yahoo.fr> wrote: >> >>> Hi Mike and everyone, >>> >>> First off, if anyone was missing my site, it is back up. I had to find >>> different >>> hosting (a bit abruptly). >>> >>> Where I was trying to link right before the site went down: >>> votingmethods.net/cond >>> works out a given (or random) scenario for Schulze, RP, or River. You >>> just have to >>> expand sections at the bottom of the result. So it could be worth a look. >>> >>> Mike wrote: >>> > Is River as easy to define &. explain as RP?. >>> >>> I see I should try to write out clearly how I suggest to understand >>> River. >>> >>> There is no "final ranking" in River. Instead every candidate begins >>> "below no one" >>> or "subordinated to no one." This is sort of a ranking but the "trees" >>> we make go >>> only one level down: you will never be able to ascend two positions from >>> a given >>> candidate. >>> >>> 1. Initially each candidate is subordinated to no one. >>> 2. Consider each pairwise defeat from strongest to weakest. >>> 3. When you consider a defeat, ask whether the loser is subordinated to >>> anyone? >>> If so: Ignore the defeat and proceed to the next. >>> If not, then ask: >>> 4. Is the defeat winner subordinated to the defeat loser? If so, ignore >>> the defeat >>> and go to the next. >>> 5. Is the defeat winner subordinated to someone else? If so, the defeat >>> loser, along >>> with everyone subordinated to them, becomes subordinated to the >>> candidate that the >>> defeat winner is subordinated to. >>> 6. Otherwise it must be that the defeat winner is subordinated to no >>> one. So here >>> the defeat loser, along with everyone subordinated to them, becomes >>> subordinated to >>> the defeat winner. >>> 7. End loop. Go to the next defeat. >>> 8. In the end, the candidates subordinated to no one are the winners. >>> >>> Alternatively instead of talking about subordination, you can say that >>> each >>> candidate has their own "bin" and starts in their own and may move to >>> another. >>> This would allow you to merge steps 4 and 5: >>> "4. The defeat loser, along with everyone *in the loser's bin*, moves to >>> whichever >>> bin the defeat winner is currently located in." >>> And if the latter bin happens to be the loser's bin, in effect nothing >>> happens. We >>> don't need a rule saying to ignore the defeat, because the bin movement >>> doesn't >>> change anything either. >>> >>> I can understand if a reader eyeballs all that and says this looks like >>> a mess and >>> it's not clearer than RP. >>> >>> But hear me out on the *ease* of it: >>> >>> 1. If you are programming River, you never actually check for a cycle, >>> whether a >>> proposed defeat would create one. And comparing to Schulze, you never >>> trace a >>> beatpath or find its strength, or (by its other algorithm) have to find >>> the Schwartz >>> set repeatedly. >>> 2. If you are solving it by hand, it would be enough to have a fridge >>> magnet for >>> each candidate, start them out in imaginary bins, and push the magnets >>> around in a >>> straightforward way to track who is subordinated to whom. >>> >>> It may be possible to define RP more concisely, but it takes some work >>> to figure out >>> what it is actually saying to do to solve it. >>> >>> Hopefully the above explains it better than I have before. >>> >>> Kevin >>> votingmethods.net >>> >> ---- >> Election-Methods mailing list - see https://electorama.com/em for list >> info >> >

On 2024-03-10 02:07, Kevin Venzke wrote: > Hi Closed Curves, Mike, > > I thought about the definition a little. Let's try this, using the "bins" concept > rather than the "subordination" one: > > 1. Initially each candidate is in their own "bin" named for themselves. > 2. Consider each pairwise defeat from strongest to weakest. > 3. For a given defeat X>Y, move everyone in bin Y to the bin that X is in. > 4. End loop. Go to the next defeat. > 5. Elect one of the candidates who remains in their original bin. > > Things to note about this: > Once Y has moved anywhere, no one will ever be in bin Y again. So we don't actually > have to throw out other defeats over Y. > If X is in Y's bin when X>Y is considered, rule 3 says that everyone in bin Y > (including X) moves to where X is, which is already Y. So nothing will happen in > this case either. > > Should be right. This concept of "bins" seems to have a pretty close connection to disjoint-set data structures, which should give River n^2 time complexity for all practical purposes. (Unless I'm mistaken, the complexity would be O(n^2 * a(n^2)), where a is the inverse Ackermann function.) -km