Partial Fix for Reweighted Range Voting (RRV)

[Toby Pereira]

It seems I'm not the first person to start a topic with this approximate title, e.g. https://groups.google.com/forum/#!topic/electionscience/ASH4YD6ButI and https://groups.google.com/forum/#!topic/electionscience/3oAWllyCwjE but this is a distinct problem I'm addressing.

One problem with RRV is how it deals with scores below the max. For simplicity, I'll assume we're using the default D'Hondt divisors. So to quote from here http://rangevoting.org/RRV.html a ballot's weight is "1/(1+SUM/MAX), where SUM is the sum of the scores that ballot gives to the winners-so-far, while MAX is the maximum allowed score". If everyone votes approval style, then it gives results that you'd expect. For example:

5 to elect, max score=10

4 voters: A=B=C=D=E=10; all other candidates = 0

1 voter: F=10; all other candidates = 0

In this case, we'd get a proportional result, such as ABCDF. But things go wrong with scores less than the max. For example:

5 to elect, max score=10

4 voters: A=B=C=D=E=7; all other candidates = 0

1 voter: F=7; all other candidates = 0

If ABCD are elected as the first five candidates, the new ballot weight for their voters would be 1/(1+28/10) = 0.263. This means that in the battle to elect the final candidate, E has a score of 4*7*0.263 = 7.37. F has a score of 7. E is elected meaning that ABCDE is the elected candidate set. This goes against what I would consider to be the proportional result. I have also discussed this problem in a couple of other threads: https://groups.google.com/forum/#!topic/electionscience/c_YRZ5ocvVg and https://groups.google.com/forum/#!topic/electionscience/f72vTfY0sC8

Another point is that while RRV elects the score winner in the single-winner case, I would argue that this is only because it is being "unnaturally" forced into doing so due to its sequential method. A non-sequential method would involve comparing each possible winning set of candidates as a whole rather than electing candidates one at a time. And based on an analogy between Forrest Simmons's non-sequential proportional approval voting and Thorvald N. Thiele's sequential version http://wiki.electorama.com/wiki/Proportional_approval_voting the non-sequential version of RRV would involve adding up satisfaction scores of voters. The satisfaction score for a voter would be HARMONIC [SUM/MAX]. The harmonic function of n is the sum of the reciprocals up to n, but generalised for non-integers as well as integers. So if we had this example:

1 voter: A=10, B=4

1 voter: A=0, B=4

The score winner would be A, and RRV would also elect A with an average of 5 compared to B's 4. However using what I would consider RRV's "natural" scoring system, the satisfaction scores would be:

A = HARMONIC [10/10] = 1

B = 2 * HARMONIC [4/10] = 1.03. http://www.wolframalpha.com/input/?i=2+*+HARMONIC+%5B4%2F10%5D

This would elect B.

So, the fix? This is the same fix that I used for the method that I developed recently. The method only worked with approval votes so I had to convert scores into approvals to get a generalised score method. It involves "splitting" voters. If a voter gives a score of 5/10 to a candidate, this converts into 0.5 voters approving the candidate and 0.5 voters not approving. This way, the total score given to each candidate is the same, but only in the form of the max score, or full approvals.

For multiple candidates, if, for example, a voter gives scores of 10/10, 8/10 and 6/10 to A, B and C respectively, this would convert to 0.6 voters approving all three candidates, 0.2 approving A and B, and 0.2 approving just A. If a candidate is approved by a part of a voter, then so will any candidate given the same or higher score. This seems to be the least arbitrary and simplest way of doing the splitting. I initially saw it as asymmetrical because I pictured each voter like a square split into tenths with the number of candidates receiving approvals going down from left to right. Using the example above (scores of 10, 8, 6), it would be 3333332211 where the left 6 tenths approve all 3 of ABC, the next 2 tenths approve of AB and the right two tenths approve of just A. But it could equally be seen like this: 1233333321, which is nice and symmetrical like a pyramid.

[Toby Pereira]

One problem with RRV is that as probably the best-known form of PR that uses score voting, it automatically gets a lot of support from people on here, despite the fact that it actually has some really bad problems - problems that have been largely ignored/unaddressed.

RRV can be seen as a generalisation in two directions of D'Hondt party list PR. The two directions being 1) it allows voters to vote for any candidates they like rather than just parties, and 2) voters can give scores rather than simply cast votes.

But the problem is that neither of these generalisations work in a reasonable way. Where voters deviate from single-party voting, it deviates away from proportionality. For example on the first generalisation:

Approval voting, proportional representation, elect 6

20 voters: A, B, C, D, E, F
10 voters: A, B, C, G

Most people asked have said that ABCDEG (or equivalent) would be the best proportional result, whereas RRV elects ABCDEF. As far as I understand, STV doesn't have this failing. RRV punishes factions for having partial agreement, whereas complete agreement or complete disagreement is OK for factions. There are other examples I could give if you want.

And as for the second generalisation (scores), I've given an example in the first post of this thread.

Here http://rangevoting.org/RRV.html it says:

"Proportionality Theorem

If some voter faction (call them the "Reds"), consisting of a fraction F (where 0≤F<1) of the voters, wants to, it is capable (regardless of what the other voters do) of electing at least ⌊(1+N)F^-⌋ red winners (assuming, of course, that at least this many red candidates run, and the total number of winners is to be N).

Specifically, it can accomplish that by voting MAX for all Reds and MIN for everybody else.

To say that again: if 37% of the voters are reds, they can assure at least about 37% red winners (up to rounding-to-integers effects). "

But this is just saying that where there is full party voting, then there will be proportionality. So it is saying that the D'Hondt part list method of PR is proportional, but it says nothing of the modifications made to turn it into RRV. These are completely unquestioned and untested. Personally I don't think they work very well at all.

[Andy Jennings]

So, the fix? This is the same fix that I used for the method that I developed recently. The method only worked with approval votes so I had to convert scores into approvals to get a generalised score method. It involves "splitting" voters. If a voter gives a score of 5/10 to a candidate, this converts into 0.5 voters approving the candidate and 0.5 voters not approving. This way, the total score given to each candidate is the same, but only in the form of the max score, or full approvals.

For multiple candidates, if, for example, a voter gives scores of 10/10, 8/10 and 6/10 to A, B and C respectively, this would convert to 0.6 voters approving all three candidates, 0.2 approving A and B, and 0.2 approving just A. If a candidate is approved by a part of a voter, then so will any candidate given the same or higher score. This seems to be the least arbitrary and simplest way of doing the splitting. I initially saw it as asymmetrical because I pictured each voter like a square split into tenths with the number of candidates receiving approvals going down from left to right. Using the example above (scores of 10, 8, 6), it would be 3333332211 where the left 6 tenths approve all 3 of ABC, the next 2 tenths approve of AB and the right two tenths approve of just A. But it could equally be seen like this: 1233333321, which is nice and symmetrical like a pyramid.

Toby,

Your fix sounds interesting.  Have you worked out what the winners would be in each of your sample scenarios here?
https://groups.google.com/forum/#!topic/electionscience/c_YRZ5ocvVg

~ Andy

[Toby Pereira]

Good question. A lot of them would be unchanged because they were approval ballots anyway, and all this fix does is turn scores into approvals. There are two criteria that I've been looking at that RRV fails. These are:

1. Independence of commonly rated candidates - This is where if a faction of voters rate a particular candidate at the same score, then the election of that candidate would not affect the further proportional allocation of candidates to any subfactions.

2. Independence of multiplication factors - If you multiply all scores by a constant, then the result would be unchanged.

It's only the second of these (Independence of multiplication factors) that has been fixed. So, for example, Election 1, which fails independence of commonly rated candidates would continue to do so.

Election 1

 

Approval voting, proportional representation, elect 6

 

20 voters: A, B, C, D, E, F

10 voters: A, B, C, G, H, I

ABCDEF would still be elected instead of the more proportional ABCDEG.

Approval elections can still exhibit a failure of independence of multiplication factors, but in a different way, as seen in Election 3.

Election 3 (mainly for those with D'Hondt leanings)

 

Approval voting, PR, elect 4

 

20 voters: A, B, C, D

10 voters: E, F, G, H

 

In this case I don't want to know your winning result (it's likely to be three from A, B, C, D and one from E, F, G, H). I want to know which you think is the better result out of A, B, C, D and A, B, E, F.

Using D'Hondt logic, ABCD should tie with ABEF. With two to elect, AE would tie with AB for the win, and this is the same proportions so they should still tie. RRV has nothing to say on ABCD v ABEF because it's a sequential system and there is no three-candidate intermediate result that could lead to both of these possibilities. However, using non-sequential satisfaction scores, ABEF would beat ABCD. This wouldn't affect the winning election result (so you could argue it doesn't matter) because ABCE would beat both anyway, but it shows that multiplying candidates can cause the same problems as multiplying scores. It's an equivalent mathematical phenomenon. The fix doesn't change anything here though because all votes are already approval votes.

Election 6

 

Score voting (max score 10), PR, elect 10

 

90 voters: A=B=C=D=E=F=G=H=I=J=4; All others=0

10 voters: K=4; All others=0

RRV elects A-J, which isn't proportional. Using the fix, it would elect A-I, K. Instead of 90 voters giving scores of 4/10 A-J, it would be as if 36 voters gave 10/10, and 4 voters gave 10/10 to K. So the fix works here.

 

 

Election 7

 

Score voting (max score 10), PR, elect 2

 

10 voters: A=10, B=4; All others=0

10 voters: B=4; All others=0

10 voters: C=10, D=4; All others=0

10 voters: D=4; All others=0

RRV would elect AC, and this would be unchanged using the fix. However, non-sequential RRV would elect BD (using HARMONIC [SUM/MAX] as the satisfaction score for voters), whereas non-sequential RRV with the scoring fix would elect AC again.

[Toby Pereira]

I might as well define my system here, since it's buried quite deep in the other thread. Starting with approval voting:

Voters cast approval ballots. If a particular candidate receives n votes, then if this candidate is elected, each voter of that candidate will get a representation of 1/n from that candidate. Each voter who did not vote for that candidate gets 0 representation from them. So the total representation received from that candidate will be 1.

A voter's total representation is the sum of the representation they receive from each elected candidate. Assuming that each elected candidate has received at least one vote, then for c candidates, the sum of the representation of all voters will be c. The arithmetic mean will always be c/v (for v voters).

Full proportionality is achieved if every voter has representation of c/v. The measure of a set of candidates is the total of the squared deviation from c/v of the voters' representation (lower being better). But also, because the variance of x is mean (x^2) - (mean x)^2, where in this case x represents voters' representation levels, (mean x)^2 will always be the same - it will be (c/v)^2 - so we can remove it from the equation.

This means that the winning set of candidates will be the set that minimises the sum of the squares of the voters' representation levels.

However, this system looks at proportionality over positive support, so in this case:

2 to elect

10 voters: A, B, C

10 voters: A, B, D

It will be indifferent between AB and CD. And in this case:

2 to elect

10 voters: A, B, C

10 voters: A, B, D

1 voter: C

1 voter: D

It will elect CD. While this may look an undesirable result in some respects, I would still argue that it is objectively more proportional. I'm yet to encounter a system that provides a trade-off between proportionality and positive support in a non-arbitrary manner. I'm not convinced one exists. But failure of Pareto and monotonicity could look bad and prone to certain strategic voting.

However, if candidates are elected sequentially, the one with most support will always be elected first anyway, and subsequent candidates must also have received reasonably high support to provide the proportional balance against this first candidate. So while arguably less proportional, the sequential method should work very well in practice. It also does, as far as I understand, collapse into Sainte-Laguë party list, if voters all vote for the candidates of one party.

The score version would work in the same way as the fix I suggested above for RRV. Because it is important for the formula that the total representation remains the same under any result (at c), if, for example, a candidate receives a single score of 1 out of 10 from one voter and nothing else, then this still has to translate into full representation from that candidate rather than 1 voter receiving a level of 0.1. This is why everything has to be translated into approval votes, and voters are "split". In this case, it translates into 0.1 of a voter fully approving the candidate.

Effectively, if scores are out of e.g. 10, then voters are split into 10. The "top" tenth of each voter approves all candidates given a score of 10. The next tenth approves candidates given 9 or 10, then 8, 9, 10 and so on down to the last tenth who approves all candidates given a score of at least 1 from that voter. This system collapses into normal score voting if just one candidate is to be elected.

In conclusion, I would argue that while the non-sequential method is the most proportional, it would still have its own problems. The sequential version would be, I would argue, strictly better than RRV. RRV's failure of independence of multiplication factors can be fixed, but it still fails independence of commonly rated candidates, which my system does not. The sequential method should also not be a problem computationally, so should be no worse than RRV in this regard.

[Jameson Quinn]

Here's some brainstorming on your theme.

You have a measure of proportionality in terms of squared deviation from your metric. A more traditional BR framework would try to maximize total approval for the slate. What if you combine the two of those via Bucklin? That is:

Voters submit ratings of candidates from 1-10

for each cutoff from 10 down to 0:

   All votes at or above cutoff are considered approvals.

   The winners under your system are calculated. (To avoid exponential computing time, we could do a sequential algorithm, and then consider whether each possible 2-way swap was an improvement, in a deterministic "last in, first out" order, until none are. That's not guaranteed to find a global optimum, but it would do pretty darn well, and it's a deterministic algorithm except for ties.)

   The total score (using original score ballots) of the winning slate is calculated

The winning slate with the highest total score is selected

... Obviously, this is massively overcomplicating things for a real-world political use. Also, my intuition is that it would reward a "picky" strategy, so maybe you'd need a rule that voters have to rate at least some fraction of the candidates above bottom-rank. (2/s? 1/sqrt[s+1]? Where s is number of seats...) 

But it does seem that there should be some way to combine your metric with the "overall score" metric. The above Bucklin-based idea is just my first stab at that. 

(note that the bucklin-based idea could be generalized/combined with your score-version of your own idea. Instead of varying a cutoff parameter and finding the level which gives the highest total score, you could take all ratings to the same power, and find the power which gives the highest total score.)

Jameson

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[Toby Pereira]

Your idea of a varying approval cut-off parameter sounds quite interesting, but if you think that you would have to restrict voters' ballots, then that's clearly a problem. Also I'm always slightly averse to measuring slates against each other by overall score, because it automatically creates bias in favour of larger factions/parties.

I certainly agree that we should be able to combine my proportionality metric with some form of overall positive support, and without creating some sort of undesirable bias, but it's just a matter of finding a satisfactory solution.

I might be being stupid here, but what do you mean by "you could take all ratings to the same power, and find the power which gives the highest total score"?

On monotonicity/Pareto, if we have the following approval ballots:

2 to elect:

50: A

1: A, B

49: B, C

(A has 51 approvals, B 50 and C 49).

Then according to my calculations, my method would elect AC, whether sequentially or non-sequentially. This is of course Pareto dominated by AB, which is what RRV would elect. So I would probably take back what I said earlier about the sequential version being *strictly* better than RRV, although I think cases like this would probably be rare and might be the price you pay for superior proportionality. B only has one more vote than C, and I'd be surprised if there were particularly egregious cases. I would still argue that my method is better overall. It would also be no good adding a Pareto condition when electing each candidate sequentially, because it wouldn't necessarily be so tidy. E.g.

5000: A

100: A, B

4900: B, C

1: C

AC is no longer Pareto dominated by AB, but it's essentially the same situation.

[Jameson Quinn]

2014-10-09 9:40 GMT-04:00 'Toby Pereira' via The Center for Election Science <electio...@googlegroups.com>:

Your idea of a varying approval cut-off parameter sounds quite interesting, but if you think that you would have to restrict voters' ballots, then that's clearly a problem.

I think the same strategic incentive exists without my idea. I haven't proved this in either case. So you can probably consider my system idea separately from my comments about strategy. 

 

Also I'm always slightly averse to measuring slates against each other by overall score, because it automatically creates bias in favour of larger factions/parties.

True. Hmm... that comment, combined with your example below, makes me think. I can't put my thoughts into coherent form just yet, but I am definitely trying to work something out. I think there's probably a coherent way to fix this. You want a system that will prefer AB in the case:

1: ABC

1: ABD

And also prefer BC in the case:

50: AB

1: BC

49: C

But that won't prefer ABC in the case:

2: ABC

1: D

Note that I say "prefer" rather than just "choose", because I don't think sequential tricks are enough.

Jameson

[Toby Pereira]

On Thursday, 9 October 2014 15:27:33 UTC+1, Jameson Quinn wrote

True. Hmm... that comment, combined with your example below, makes me think. I can't put my thoughts into coherent form just yet, but I am definitely trying to work something out. I think there's probably a coherent way to fix this. You want a system that will prefer AB in the case:

1: ABC

1: ABD

And also prefer BC in the case:

50: AB

1: BC

49: C

But that won't prefer ABC in the case:

2: ABC

1: D

Note that I say "prefer" rather than just "choose", because I don't think sequential tricks are enough.

Jameson

Yes. But also in this case with two to elect:

10: AC

10: BC

It would prefer AC or BC over AB. And because of the amount of advantage given to one of the factions, it would arguably not be desirable, even though C has unanimous support. But yes, it would be nice to have this system exist, and then we can debate its pros and cons once it does.

As we discussed on the elections methods mailing list, it would be the ideal proportional method for meals if you could only have one meal of each type. So with two meals to have:

10: Pizza, fry-up

10: Pizza, curry

Nothing is gained by bringing other people down to your level, so you'd have pizza on one of the days and either fry-up or curry on the other. Obviously the way Forest Simmons's PAV method works (which is also the approval version of RRV), higher support is preferred, but independence of commonly rated candidates is still a sticking point for me. I was after a fix for this for years before I even considered dropping the monotonicity/Pareto requirement. I've spent longer not finding this method than I took to find the method I did find.

[Toby Pereira]

Something occurred to me about how this monotonic system would have to work. As it is if you have this:

Approval voting, 2 to elect

30: A, B

10: C

There should be a tie between AB and AC. But if you have:

Approval voting, 4 to elect

30: A, B, C, D, E

30: A, B, C, D, F

20: G, H

Although you can have a proportional result (e.g. ABCG), there should still be a non-winning tie between ABCD and ABGH. But also there should still be a tie between ABEF and ABGH (after all C and D might not exist and you'd want this tie then). But ABCD has to beat ABEF, so this leads to a contradiction. This is big.

So, what do we do? If you use Simmons PAV (with Sainte-Laguë divisors), then you will get a tie between AB and AC in the first example of this post. But you won't get a tie between ABCD and ABGH even though it's exactly the same apart from multiplying everything up. But it still "works" in cases like this because it still elects the best winner. It's just that things that would have tied for the win in one election could only tie for a non-win when things are multiplied up so it escapes giving the wrong result.

So the point is that I don't think we can expect a monotonic system to correctly order all slates of candidates. We have to be happy with it finding the winner. So at the moment I'm trying to find ways of modifying the Simmons method to obey independence of commonly rated candidates rather than modifying my method to pass monotonicity.

[Toby Pereira]

On further reflection, I'm going to take back what I said about all those results having to be equal (specifically between ABEF and ABGH).

But the question we need to answer is: when two results are proportionally equal, by what measure to we distinguish them? For example with two to elect and approval voting:

10: A, B, C

10: A, B, D

It's quite easy to see that while AB and CD might measure as equally proportional, AB is the stronger result. But not all cases are this easy, so we need an exact way of measuring it. If you have:

2 to elect, approval voting

30: AB

10: C

Then AB and AC are equally proportional, but I would argue that they are exactly equal in all relevant senses here. But how are we measuring it? One more example:

4 to elect, approval voting

30: A, B, G, H, I, J

10: C, K, L

30: D, E, G, H, I, J

10: F, K, L

Ignore that these might not be winning results, but there is a proportional tie between ABDE, ABDF, ACDE, ACDF, GHIJ, GHKL. But we can see that GHIJ and GHKL are stronger results because everyone's number of seats is doubled by the combining of factions. But other than this intuitive sense in these simple cases, what is the actual measure?

[Toby Pereira]

One possible solution to this is to convert levels of proportionality into scores, and then "split" each result in order to make the highest possible score. For example:

2 to elect, approval voting

10 voters: A, B, C

10 voters: A, B, D

As discussed before, AB and CD are equally proportional. Because they're both perfectly proportional by my measure, we'll say they have a proportionality score of 1. But the AB result can be split. Looking at the ballots with just A and B, we have

10: A, B

10: A, B

(or just 20: A, B)

But this can be split into:

10: A

10: B

plus

10: B

10: A

Each of which would have a proportionality score of 1, so this would give a total score of 2. The idea is that every vote still gets used exactly once and we find the most efficient way of doing it. Obviously this could be a big computational problem, but I'm not going to worry about that for now. Also, other than a score of 1, it's not necessarily obvious how to score each level of proportionality. But this would clearly be monotonic because adding in an extra vote for a candidate could only ever increase the total score for the slates with this candidate in, and leave the score for all other slates unchanged.

If we have:

30 voters: AB

10 voters: C

Then AB and AC are equally proportional. AC results in every voter getting exactly one seat each (out of two). Then if we have:

20 voters: A, B, D

10 voters: A, B, E

10 voters: C, E

We can see that the voters are the same with respect to A, B and C as in the previous example, so AB still equals AC. But we also have DE which is a perfectly proportional result. This, like AC, gives every voter one seat each. And using Simmons PAV, it's easy to see that AC and DE would be rated the same, even though DE is clearly the more proportional result. But this makes sense with our earlier food example because everyone has one type of food that they like and it's not shared out.

But AC can't be obviously split into more than one part that add up to make DE. So I'm not sure this system works here. However, it might be more fruitful to find for each result an equivalent "flat" result. For example, AB is proportionally equivalent to AC, and AC gives each voter 50% of the seats each, so both AB and AC could be seen as having a score of 0.5. DE also gives everyone one seat each so has a score of 0.5. AE would actually be the best of the lot because everyone who voted for C also voted for E and some people voted for E and not C, so AE beats AC and everything that was level with AC. (Simmons PAV gives AE the best score).

But the problem here is to find an "equivalent flat result" for each actual result.

[Toby Pereira]

I was doing some searching, and it seems that the criterion I've named independence of commonly rated candidates has been around for over 100 years, and was used by Lars Edvard Phragmén and Gustav Cassel to criticise Thorvald Thiele's sequential proportional approval voting method (which is the sequential version of Forest Simmons's method).

Phragmén developed his own system to counter this problem.

https://groups.yahoo.com/neo/groups/election-methods-list/conversations/topics/10068

https://groups.yahoo.com/neo/groups/election-methods-list/conversations/topics/10276

I haven't exactly got my head round how it works, but then I found this as well:

https://groups.yahoo.com/neo/groups/election-methods-list/conversations/messages/12468

Bjarke Dahl Ebert described it in a simpler manner (assuming it's correct) in 2003. Where I would minimise the sum of the squares of the voters' representation scores, Phragmén would apparently minimise the maximum representation score (L-infinity norm). Ebert then suggested some modifications to make it even better and has ended up with the same method as me! He seems less sure of it because it doesn't collapse into D'Hondt unlike Phragmén's method - but I see this as a good thing.

This is all very interesting, but when I first came across this I imagined that Phragmén's method would be a modification of the Thiele method to make it pass independence of commonly rated candidates. But actually it seems that it would, like mine, fail monotonicity as well. So it gains nothing in terms of this project.

One other thing - in post 12 of this thread:

http://www.revleft.com/vb/voting-ratings-t143429/index.html?s=efdb1dc1efca4ca889fe1d7d879ac089&amp;t=143429

Someone called Kotze describes Phragmén's method and explains how it could also be used with a score ballot, and he suggests the same score generalisation as I do! Turns out I've come up with nothing new after all!

[Clay Shentrup]

On Sunday, October 26, 2014 5:15:24 PM UTC-7, Toby Pereira wrote:

I was doing some searching, and it seems that the criterion I've named independence of commonly rated candidates has been around for over 100 years, and was used by Lars Edvard Phragmén and Gustav Cassel to criticise Thorvald Thiele's sequential proportional approval voting method (which is the sequential version of Forest Simmons's method).

This is nuts. You can picture these people generations ago having the same conversations we're having now. Could find in any case.

[Toby Pereira]

Yeah, it's weird. In a way it shows how little we've progressed in some areas! But I think it also goes to show how easily this stuff can get lost. It was obviously rediscovered about 10 years ago, which is how I found those posts on it, but then it seems to have been forgotten about again. I might set up an account on the Electorama wiki (assuming it's open to that) and put some stuff on there. I don't think there is a very good single place to find out about voting systems. The Wikipedia is good for ones that are deemed to be "notable", but the Electorama wiki has the potential to be much better. But it seems to be largely gathering dust, and isn't very navigable.

[Toby Pereira]

I was thinking more about how to base a method around "splitting" results. If you have the following example:

2 to elect, approval voting

3: A, B, C

3: A, B, D

1: C

1: D

My method would say CD is the most proportional result. But because it's not that "strong" a result, the Sainte-Laguë version of Simmons PAV would say that CD is equal to AB. This looks on the face of it the sort of result that seems acceptable if you want a system that's proportional up to the point where it would lose monotonicity. If we trust it here, then using my splitting method, we would have CD having a proportionality of 1. And we'd also want the proportionality scores for AB to add up to 1. So looking at just the AB votes:

3: A

3: B

2: -

+

3: B

3: A

2: -

=1

Since the AB result can be split into two, then it's actual proportionality level should be 0.5, so that its score is 1.

Previously it had been suggested that a reasonable proportionality score would be mean (x^2) / (mean x)^2, where x is the voters' representation scores. Perfect proportionality would score at 1. But this AB result would have a score of 0.75. So the formula would need further tweaking to convert the 0.75 into 0.5. Similarly, another "equilibrium" result for Sainte-Laguë PAV is:

3 to elect

15: A, B, C, D

15: A, B, C, E

15: A, B, C, F

8: D

8: E

8: F

ABC is equal to DEF. Because the ABC result can be split into three this time, we'd want its proportionality level to be 1/3. Its mean (x^2) - (mean x)^2 level would be 15/23. 15/23 is 1/(1 + 1/3 + 1/5).

And similarly, we would want 1/(1 + 1/3 + 1/5 + 1/7) (which is 105/176) to convert to 1/4 and so on. So we have:

3/4 -> 1/2

15/23 -> 1/3

105/176 -> 1/4

I could fiddle around to come up with a general formula, but it would be nasty and would I think have to involve the inverse of the harmonic number function, and not something you'd really want in a voting system that you're trying to keep as simple as possible. Also, it would need to be shown that these are the correct conversion numbers in every case, and I'm not convinced that this is the correct direction to be going in. I generally don't trust the divisors here because the Simmons PAV method can be shown to give weird results when you multiply things up and try to generalise. Also that 15/8 ratio above seems clunky and arbitrary and I wonder if its a case of the system being pushed further than is reasonable. An example of PAV breaking down:

Approval voting, 2 to elect

3: AB

1: C

Sainte-Laguë PAV says that AB and AC are equal. But if you have:

4 to elect

3: ABCD

1: EF

Then ABCD and ABEF are not equal. They aren't winning results anyway (ABCE, for example, beats both), but it's a reason I don't trust it when you try to generalise.

And just finally, with this example from earlier:

20 voters: A, B, D

10 voters: A, B, E

10 voters: C, E

I suggested that AC and DE would be considered equal in the food example because everyone gets one type of food they like and one they don't. But in a political election, I would always consider DE superior, and the splitting method (as far as it goes) would agree with me here (although Simmons PAV rates them the same). So it could be that we're looking for more than one monotonic method, to cover all types of election.

[Warren D Smith]

On 10/26/14, 'Toby Pereira' via The Center for Election Science

<electio...@googlegroups.com> wrote:
> I was doing some searching, and it seems that the criterion I've named
> independence of commonly rated candidates has been around for over 100
> years, and was used by Lars Edvard Phragmén and Gustav Cassel to criticise
> Thorvald Thiele's sequential proportional approval voting method (which is
> the sequential version of Forest Simmons's method).

--Thorvald N. Thiele and Lars Edvard Phragmen were
both noted mathematicians at that time in Sweden.
(Well, by "noted" I mean "I'd heard of some of their non-voting-related math.")
I'd never heard of Karl Gustav Cassel before, but wikipedia has an
article on him:
http://en.wikipedia.org/wiki/Gustav_Cassel
saying he was "a founding member of the Swedish school of economics"
and "his authority was second only to that of Lord JM Keynes."

I would guess that they all were stimulated to examine voting by the
democratization of Sweden in the early part of the 20th century.

As Toby said,

https://groups.yahoo.com/neo/groups/election-methods-list/conversations/messages/12468

is a pretty clear post by Bjarke Dahl Ebert allegedly
describing Phragmen's multiwinner voting method and some suggested changes
(hopefully improvements). He does not cite any original work by Phragmen but
another post cited
E. Phragmen: "Till frågan om en proportionell valmetod" [On the
Question of a
Proportional Election Method], Statsvetenskaplig Tidskrift (StvT,) 2,7
(1899) 87-95.
It seems reasonable to me at least on the surface.
I'd like to see some theorems proven about it -- is there a global optimum,
which their "greedy" approach finds? What is the relation with
http://RangeVoting.org/MonroeMW.html?

[Toby Pereira]

On Monday, 27 October 2014 21:00:57 UTC, Warren D. Smith (CRV cofounder, http://RangeVoting.org) wrote:

As Toby said,

https://groups.yahoo.com/neo/groups/election-methods-list/conversations/messages/12468

is a pretty clear post by Bjarke Dahl Ebert allegedly
describing Phragmen's multiwinner voting method and some suggested changes
(hopefully improvements).  He does not cite any original work by Phragmen but
another post cited
    E. Phragmen: "Till frågan om en proportionell valmetod" [On the
Question of a
Proportional Election Method], Statsvetenskaplig Tidskrift (StvT,) 2,7
(1899) 87-95.
It seems reasonable to me at least on the surface.
I'd like to see some theorems proven about it -- is there a global optimum,
which their "greedy" approach finds?  What is the relation with
   http://RangeVoting.org/MonroeMW.html?

I'd need to go through Phragmén's method again to fully understand it, because it would seem strange to me if it's just a case of minimising the L-infinity norm - so just a case of minimising what I would call the "representation score"* of the voter with the highest such score. Because it's a sequential method, we could if we wanted simply declare five candidates to be elected and then elect the sixth properly using Phragmén's method. If one voter in particular had a much better deal than anyone else from these first five, it could be that whoever is elected as the sixth candidate would not change the highest individual voter's representation score (this one voter might not have voted for anyone not among these five candidates). This would mean that if Ebert's interpretation of the method is correct, we could contrive cases where Phragmén's method would be indifferent between all candidates for the final position.

If Ebert is correct, then there would be a global optimum, and it would be the result that minimises the highest individual voter's representation score.

With Monroe's method, I think I understood it best from Andy Jennings's post here - http://permalink.gmane.org/gmane.politics.election-methods/19720

"If you want a clustering PR method, then I would highly recommend Monroe.  In Monroe, the score of each slate is equal to the sum of each voter's score of his assigned candidate.  I think that, for a given slate, finding the optimal way of assigning the voters to the winners is polynomial. (See Warren's page for more information: http://rangevoting.org/MonroeMW.html)"

Because, as far as I understand, it optimally assigns each voter to a candidate (with every candidate having an equal number of voters), it makes no difference what each voter thinks of anyone other than their own candidate. In fact, Warren says as much as a criticism of the method. This seems very different from Phragmén's method, which seems to take into account all candidates that a voter has voted for. Because of that I don't think "full proportional representation" is particularly full.

The method that Ebert described, which is identical the method that I have also independently discovered, can be argued to be the most proportional method for a reasonable definition of proportionality. Its problem, however, is that it fails monotonicity and Pareto as a sacrifice for proportionality. After Ebert's description, there seem to be no follow-up posts. He seemed less excited by it than me anyway, and it was allowed to get forgotten. It's possible that others have discovered it too, because it's not overly complex.

*Just to clarify the representation score - if 1/v voters vote for a particular elected candidate, then everyone who has voted for that candidate gets a representation level of 1/v from that candidate. Everyone else gets 0. A voter's representation score is the total of the representation levels they get from the elected candidates.

[Toby Pereira]

On Monday, 27 October 2014 19:37:35 UTC, Toby Pereira wrote:

 

Previously it had been suggested that a reasonable proportionality score would be mean (x^2) / (mean x)^2, where x is the voters' representation scores. Perfect proportionality would score at 1. But this AB result would have a score of 0.75. So the formula would need further tweaking to convert the 0.75 into 0.5. Similarly, another "equilibrium" result for Sainte-Laguë PAV is:

3 to elect

15: A, B, C, D

15: A, B, C, E

15: A, B, C, F

8: D

8: E

8: F

ABC is equal to DEF. Because the ABC result can be split into three this time, we'd want its proportionality level to be 1/3. Its mean (x^2) - (mean x)^2 level would be 15/23. 15/23 is 1/(1 + 1/3 + 1/5).

And similarly, we would want 1/(1 + 1/3 + 1/5 + 1/7) (which is 105/176) to convert to 1/4 and so on. So we have:

3/4 -> 1/2

15/23 -> 1/3

105/176 -> 1/4

I could fiddle around to come up with a general formula, but it would be nasty and would I think have to involve the inverse of the harmonic number function, and not something you'd really want in a voting system that you're trying to keep as simple as possible. Also, it would need to be shown that these are the correct conversion numbers in every case, and I'm not convinced that this is the correct direction to be going in. I generally don't trust the divisors here because the Simmons PAV method can be shown to give weird results when you multiply things up and try to generalise. Also that 15/8 ratio above seems clunky and arbitrary and I wonder if its a case of the system being pushed further than is reasonable.

There's a much better way of dealing with this. When I first looked at finding a monotonic proportional score/approval system, I found that squared dissatisfaction scores worked in cases where there were two factions. So, for example:

Approval voting, 2 to elect

3 voters: A, B

1 voter: C

For the result AB, 3 voters have 0 dissatisfaction and 1 voter has 2. So we have 3 * 0^2 + 1 * 2^2 = 4

For AC, it's 4 * 1^2 = 4.

These two results are equivalent. In this case it means that 3/4 of the voters getting everything (1) and 1/4 getting nothing (0) is equivalent to everyone getting 1/2. This translates well into the examples I've been using here:

Approval voting, 2 to elect:

3 voters: A, B, C

3 voters: A, B, D

1 voter: C

1 voter: D

For AB we have 6 * 0^2 + 2 * 2^2 = 8

For CD we have 8 * 1^2 = 8

The next example up gives us:

Approval voting, 3 to elect

5 voters: A, B, C, D

5 voters: A, B, C, E

5 voters: A, B, C, F

4 voters: D

4 voters: E

4 voters: F

ABC - 15 * 0^2 + 12 * 3^2 = 108

DEF - 27 * 2^2 = 108

So here we have a 5/4 ratio rather than 15/8 that I found with Sainte-Laguë PAV. And instead of the raw proportionality level of 15/23 translating into an additive score of 1/3, we have a raw proportionality level of 5/9 translating into a score of 1/3. If we have a proportionality level of x and want to find the additive score y, we have y = 1 - sqrt (1-x), which doesn't involve anything horrible like the inverse harmonic number function.

I think there is a method in here now. Add up the squares of the representation scores of the voters as you would anyway in the method discovered independently by Bjarke Dahl Ebert and me. Divide the total you'd get if there was perfect proportionality by this number, and you get a proportionality level between 0 and 1. If this number is x, then you can turn it into an additive score with a formula: score = 1 - sqrt (1-x).

Then you find the optimum "split" of the vote, which gives the highest total score. For example:

20: AB

Can be split into two lots of

10: A

10: B

Every vote must be used exactly once. If this method does what I want it to, then it would provide the optimum blend of proportionality and monotonicity. But if we have a look at the food example again:

Approval voting, 2 to elect

20 voters: Pizza, Curry

10 voters: Pizza, Pasta

10 voters: Fry-up, Pasta

There are two meals to be had and each is available only once. Ignoring Pizza/Pasta, Curry/Pasta is clearly more proportional than Pizza/Fry-up and would be the preferred result in either version of my method. However, in this case both simply result in every person having one choice they like and one they don't, so it makes sense to say they are equal. So it also makes sense to carry on the search for a third type of method.

[Toby Pereira]

Just to tidy up the definition of the new method:

Voters cast approval ballots (see top post of the thread for the score conversion). If a particular candidate receives n votes, then if this candidate is elected, each voter of that candidate will get a representation of 1/n from that candidate. Each voter who did not vote for that candidate gets 0 representation from them. So the total representation received from each elected candidate will be 1.

A voter's total representation is the sum of the representation they receive from each elected candidate. Assuming that each elected candidate has received at least one vote (if not, it will break the system with division by zero), then for c candidates, the sum of the representation of all voters will be c. The arithmetic mean will always be c/v (for v voters).

Full proportionality is achieved if every voter has representation of c/v. If we take a voter's representation score to be x, then the overall proportionality score of the result is:

1 - (sqrt (1 - (mean (x^2) / (mean x)^2)))

This will always be between 0 and 1. For a given slate of candidates, the set of ballots can then be split into several ballot subsets, with the same total votes. For example, if a particular voter cast this ballot:

Voter 1: A, B, C, D

Then this could be split across several "subelections" as long as each vote is used exactly once. For example:

Subelection 1: A

Subelection 2: B, C

Subelection 3: -

Subelection 4: D

Every voter would have their votes split in some manner across these subelections. For a given result, each subelection would have its own proportionality score, which are added together. The optimum split for a slate of candidates is the one that gives the highest total of the proportionality scores. The winning slate of candidates is the one that has the highest possible total of the proportionality scores. For example:

Approval voting, 2 to elect

10 voters: A, B, C

10 voters: A, B, D

AB and BC are both perfectly proportional so each have a proportionality score of 1. However, AB can be split into:

10 voters: A

10 voters: B

+

10 voters: B

10 voters: A

Each of these ballot sets would have a proportionality score of 1 under the result AB, so we can add them together and say that AB has a total score of 2 because that is its score under the optimum split. This beats CD with its own score of 1.

[Toby Pereira]

On Monday, 27 October 2014 21:00:57 UTC, Warren D. Smith (CRV cofounder, http://RangeVoting.org) wrote:

https://groups.yahoo.com/neo/groups/election-methods-list/conversations/messages/12468

is a pretty clear post by Bjarke Dahl Ebert allegedly
describing Phragmen's multiwinner voting method and some suggested changes
(hopefully improvements).  He does not cite any original work by Phragmen but
another post cited
    E. Phragmen: "Till frågan om en proportionell valmetod" [On the
Question of a
Proportional Election Method], Statsvetenskaplig Tidskrift (StvT,) 2,7
(1899) 87-95.
It seems reasonable to me at least on the surface.
I'd like to see some theorems proven about it -- is there a global optimum,
which their "greedy" approach finds?  What is the relation with
   http://RangeVoting.org/MonroeMW.html?

I had another look at the description that Olli Salmi gave of Phragmén's method, and as far as I can see it's not exactly the same as Bjarke Dahl Ebert's interpretation.

According to Salmi's description of this sequential method, the next candidate to be elected will be the one that minimises the average "load" (what I call the representation score) of the voters that vote for the next elected candidate. A global optimum for this, I would suggest, is to minimise the average load for voters of the elected candidate whose voters have the highest average load. Ebert's interpretation is to minimise the load of the individual voter with the highest load, so minimises the maximum load.

In either case, it is clear that any global optimum of Phragmén's would still fail monotonicty. For example:

Approval voting, 2 to elect

10: A, B, C

10: A, B, D

It would be indifferent between AB and CD, just as Ebert's modification (the same as my method) would. And:

Approval voting, 2 to elect

10: A, B

10: A, C

It would elect BC, ignoring the unanimously supported A.

I think Salmi was wrong in his assumption that when seeing which candidate to elect next, you could save time by ignoring certain candidates that were dominated by other candidates. For my/Ebert's method, I found a Pareto failure even when electing sequentially, so I will test that here too.

Approval voting, 2 to elect:

50: A

1: A, B

49: B, C

Overall, we have A=51, B=50, C=49.

A is elected first with the most approvals. Looking at the loads, we have

50: A - 0.980

1: A, B - 0.0196

49: B, C - 0

If we elect B next, the total load on the B voters will be 1+0.0196 = 1.0196. The average load will be 1.0196/50 = 0.02039.

If we elect C next, the total load on the C voters will be 1. The average load will be 1/49 = 0.02041.

Under Salmi's interpretation, AB would be elected, so it wouldn't fail Pareto in this case but that doesn't make it impossible (I will try other examples in the absence of any proof).

However, under Ebert's interpretation, if AC is elected, then the highest individual load for a voter is 1/49 (the 49 C voters all have this load). If AB is elected, then the single AB voter has a load of 1/50 + 1/51, which is more than 1/49, so under Ebert's interpretation, AC would be elected.

In any case, I would argue that Phragmén's method is still less proportional overall than my/Ebert's method, so it's not a method I would advocate the use of in any particular situation. If I can demonstrate a Pareto failure, then all the more so.

[Toby Pereira]

Phragmén does fail Pareto.

34: A

33: B, C

32: D, E

1: A, C, E

Overall we have A=35, B=33, B=34, D=32, E=33.

A has the most approvals so is elected first. Looking at the loads:

34: A - 0.9714 (34/35)

33: B, C - 0

32: D, E - 0

1: A, C, E - 0.02857 (1/35)

If B is elected next, the average load on B voters is 1/33 = 0.03030

If C is elected next, the average load on C voters is (1/35 + 1) / 34 = 0.03025

If D is elected next, the average load on D voters is 1/32 = 0.03125

If E is elected next, the average load on E voters is (1/35 + 1) / 33 = 0.03117

C voters have the lowest average load, so C is elected second. Now the loads are:

34: A - 0.9714 (34/35)

33: B, C - 0.9706 (33/34)

32: D, E - 0

1: A, C, E - 0.05798 (1/35 + 1/34)

If B is elected next, the average load on B voters is (33/34 + 1) / 33 = 0.05971

If D is elected next, the average load on D voters is 1/32 = 0.03125

If E is elected next, the average load on E voters is (1/35 + 1/34 + 1) / 33 = 0.03206

D voters have the lowest average load, so D is elected next. However, D is Pareto dominated by E, meaning that Phragmén's method fails the Pareto criterion.

Having done this, it took some quite contrived cases to get either Phragmén's method or the sequential version of my/Ebert's method to fail, and I don't think either method would show up too many problems if used in real elections. In real life elections with many candidates, they would likely be used sequentially anyway, so the monotonicity/Pareto problems that they might face when used non-sequentially aren't relevant. It would likely add significant complexity to get a reasonable proportional method that would pass monotonicity and Pareto (e.g. my splitting method is complex and still doesn't fully work in its current form), and given that the results are likely to be similar to Phragmén's method or mine/Ebert's when used sequentially, it might not be worth the added complexity. But I will carry on looking at alternative methods.

[Toby Pereira]

A few times it's been suggested that the non-monotonic proportionality score could be combined with the overall rating for a set of candidates to make a proportional and monotonic system. I've resisted this idea because by putting overall score onto the proportionality, you're automatically favouring larger factions. However, the raw proportionality measure that I've been using obeys a Sainte-Laguë proportionality, and it occurred to me that by multiplying by total score, it might become a D'Hondt version. With this example:

2 to elect, approval voting

30: A, B

10: C

AB and AC are equally proportional. But with the AB score, it would be equally proportional if 15 voters voted solely for A and 15 solely for B even if it would mean that the result would have less overall support. And using my "splitting" logic from earlier, it would seem that perhaps AB (with the original ballot) is better than AC using a monotonic measure. While I haven't yet generated a proof, it seems that multiplying proportionality by support (proportion of maximum approval) - Gabriel Bodeen suggested doing this - does give D'Hondt type results in simple cases. For example:

2 to elect, approval voting

2: A, B

1: C

AB would tie with AC

3 to elect, approval voting

3: A, B, C

1: D

ABC would equal ABD. It seems that generally:

x to elect, approval voting

x: A1, A2...Ax

1: B

A1, A2...Ax equals A1, A2...Ax-1, B as would fit D'Hondt.

I haven't proved that result, but it's worked for everything I've tried and if it's true it shouldn't be too hard to generate a proof. So I thought that this might be the method I've been after, and I was interested to see if it obeyed independence of commonly rated candidates. So I tried this:

2 to elect, approval voting

2: A, B, C

1: A, D

A is approved by everyone so should be ignored. Then there should be a D'Hondt tie between BC and BD for the other two seats. So I multiplied the 0-1 proportionality scores by the 0-1 approval scores for both these results. For ABC I got 0.63636 (proportionality 0.8181, approval 0.7778). For ABD I got 0.63158 (proportionality 0.9474, approval 0.6667). It's really close, but failure is failure, and it's not due to rounding. It's not like the Simmons method and RRV where it fails completely, but any failure makes it unacceptable. I'm not sure if there's a way to fix it. It's not as intuitively understandable as the proportionality measure on its own, so it's not obvious to me how it's going to behave in any given situation, which makes it harder to work with and know how to make any changes. Its D'Hondt-type results were a discovery for me when I did the calculations, whereas the Sainte-Laguë results from the pure proportionality measure are fairly obvious from the way it was developed.

[William Waugh]

I don't see why you suppose that some of the voters don't use the full range.  Isn't it reasonable to expect voters would not voluntarily give up political power?

On Monday, September 8, 2014 at 5:17:49 PM UTC-4, Toby Pereira wrote https://groups.google.com/d/msg/electionscience/BURkyIxZaBM/GdAcH2z7XkEJ

[Bruce Gilson]

On Tue, Jan 20, 2015 at 1:12 AM, William Waugh <google_wil...@spamgourmet.com> wrote:

I don't see why you suppose that some of the voters don't use the full range.  Isn't it reasonable to expect voters would not

​​

voluntarily give up political power?

​People aren't always so knowledgeable ​about the voting systems. A voter faced with two candidates he likes, but one better than another, might give one a 10 and one a 9 on a 1-10 scale even though he would have 9 times

 

​the power using 10 and 1, because he doesn't really realize he is "

​

voluntarily giv

​[ing]

 up political power

​.​

"​

[Toby Pereira]

But also I would want a voting system to "work" regardless of how rational voters actually are. I could change the example to a less irrational one anyway. Instead of this example:

5 to elect, max score=10

4 voters: A=B=C=D=E=7; all other candidates = 0

1 voter: F=7; all other candidates = 0

I could change it so that each faction does have at least one candidate that they give a top score to. The larger faction could have four candidates they give a top score to and the smaller faction has one candidate they give a top score to.

4 voters: A=B=C=D=E=7; G=H=I=J=10; all other candidates = 0

1 voter: F=7; K=10; all other candidates = 0

If five candidates are elected, then it would be these five. But then if more than five candidates are elected, it will move onto the 7s, and these will then be elected in a less than proportional manner, as already argued. But just to play it out, say there are 10 to be elected. Proportionally, it should be 8 to the larger faction and 2 to the smaller faction. If the first 9 elected are all the 10s and A to D, we would expect F to be preferred to E, proportionally speaking.

Using the 1/(1+SUM/MAX) ballot weight formula, the larger faction's ballots are weighted at 1/(1+68/10) = 0.128.

The smaller faction's ballots are weighted at 1/(1+10/10) = 0.5.

This gives E a score of 4*7*0.128 = 3.590.

F has a score of 1*7*0.5 = 3.5.

E is (disproportionally) elected.