1. Do an n-1 out of n election.
2. If n-1 is equal to the number of seats, you're done and that's the 

First: Let's say that method B is a more decisive version of method A if you can get B by breaking ties in the outcomes produced by A. (I.e. B's outcome never ranks X ahead of Y whn A ranks Y ahead of X, but sometimes ranks X ahead of Y when A ranks X equal to Y.) Then it's possible for B to be summable even when A isn't. So tiebreaking can make matters *easier*. (This unfortunately makes election method design harder.) Second: Suppose you have some black box method that only does "n-1 out of n" multiwinner elections, i.e. where the number of seats is one less than the number of candidates. Suppose this method is Droop proportional. Then the following multiwinner method is Droop proportional in general and extends the method to any-seat elections: 1. Do an n-1 out of n election. 2. If n-1 is equal to the number of seats, you're done and that's the outcome. 3. Otherwise, eliminate whoever was not elected and go to 1. My proof idea is like this: Suppose that we're trying to elect k seats and there are n candidates, and |V| voters. Then the Droop quota for k seats is |V|/(k+1). So suppose that the Droop proportionality criterion states that q candidates from some coalition must be elected because q|V|/(k+1) voters ranked them first. Then since the Droop quota for n-1 seats is lower than that for k seats, at least q candidates from the coalition must also be elected in the (n-1) seat election. Hence the nth candidate (the one who's not elected when the (n-1) seat election is performed) can't be any of the q. So eliminating him can't violate any future Droop constraints. The rest follows pretty straightforwardly from induction. So in a sense, the hard part of a proportional multiwinner method is the (n-1) out of n election. Solve that and you can get Droop proportionality in general -- although it would probably be neither monotone nor summable. The funny/interesting part is that no actual multiwinner election method I know of is constructed this way. Perhaps something notable could be found in the space of election methods that look like this? -km

     1. Do an n-1 out of n election.
     2. If n-1 is equal to the number of seats, you're done and that's

Cool ... I think you're onto something good! On Mon, Feb 20, 2023, 6:08 AM Kristofer Munsterhjelm <km_elmet@t-online.de> wrote: > First: Let's say that method B is a more decisive version of method A if > you can get B by breaking ties in the outcomes produced by A. (I.e. B's > outcome never ranks X ahead of Y whn A ranks Y ahead of X, but sometimes > ranks X ahead of Y when A ranks X equal to Y.) > > Then it's possible for B to be summable even when A isn't. So > tiebreaking can make matters *easier*. (This unfortunately makes > election method design harder.) > > Second: Suppose you have some black box method that only does "n-1 out > of n" multiwinner elections, i.e. where the number of seats is one less > than the number of candidates. Suppose this method is Droop > proportional. Then the following multiwinner method is Droop > proportional in general and extends the method to any-seat elections: > > 1. Do an n-1 out of n election. > 2. If n-1 is equal to the number of seats, you're done and that's > the > outcome. > 3. Otherwise, eliminate whoever was not elected and go to 1. > > My proof idea is like this: Suppose that we're trying to elect k seats > and there are n candidates, and |V| voters. Then the Droop quota for k > seats is |V|/(k+1). So suppose that the Droop proportionality criterion > states that q candidates from some coalition must be elected because > q|V|/(k+1) voters ranked them first. Then since the Droop quota for n-1 > seats is lower than that for k seats, at least q candidates from the > coalition must also be elected in the (n-1) seat election. Hence the nth > candidate (the one who's not elected when the (n-1) seat election is > performed) can't be any of the q. So eliminating him can't violate any > future Droop constraints. > > The rest follows pretty straightforwardly from induction. > > So in a sense, the hard part of a proportional multiwinner method is the > (n-1) out of n election. Solve that and you can get Droop > proportionality in general -- although it would probably be neither > monotone nor summable. > > The funny/interesting part is that no actual multiwinner election method > I know of is constructed this way. Perhaps something notable could be > found in the space of election methods that look like this? > > -km > ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

On 2/21/23 02:24, Forest Simmons wrote: > Cool ... I think you're onto something good! Thanks :-) Observations for three candidates: Let a candidate who is barred from being included in an n-1-seat council due to Droop criterion constraints be called "unfortunate". For three candidates, the Droop quota is |V|/3. The possible exclusion scenarios are: If A has more than |V|/3 first preferences, A is not unfortunate. Ditto B and C. If there exists some solid coalition {A, B} supported by more than two Droop quotas, then C is unfortunate. In other words, if C's last preference count is greater than 2|V|/3, then C is unfortunate. If a solid coalition of two candidates is supported by more than one Droop quota, but not more than two, then we don't need to care about it because we can never eliminate both members of the solid coalition. Even in a case where we have a Droop quota voting B>A and another Droop quota voting C>A, we don't necessarily need to preserve A (although it may seem natural to do so). Hence A is unfortunate if fpB > |V|/3 and fpC > |V|/3 or lpA > 2|V|/3. Note that an unfortunate candidate may not necessarily exist. E.g. 34: A>B>C 33: B>C>A 32: C>A>B So if we were to create a method that finds a loser candidate, every unfortunate candidate should lead to that unfortunate being kicked out, but it should also kick out a candidate that's not unfortunate in the cases when such doesn't exist. Some people have suggested using IRV for this purpose, but the criteria for unfortunates show that it doesn't work. For instance: 1: A>B>C 3: C>B>A B has no first preferences and is eliminated first, but A is unfortunate because more than two Droop quotas support {B, C}. Two interesting questions I don't know the answer to: 1. Is it possible to create a method that in the three-candidate case always evicts unfortunates, is decisive, and only uses pairwise information? 2. For higher number of candidates, do we only have to look at "a single Droop quota supports this candidate" and "more than n-1 Droop quotas support everybody but this candidate" criteria? Or can size-k solid coalitions interact in complex ways to protect certain candidates from elimination? -km