Possible improvement to PAMSAC (gets rid of CFAT)

What about using ABC + (AB + BC + CA)/3 + (A + B + C)/3 + 1?

I just figured out that AB + BC + CA = ABC + A + B + C.

How does it generalize with 5 winners? 5 is the typical winner set size.

AB = AB + 0*(A+B)
AB + BC + CA = ABC + 1*(A+B+C)
AB + BC + CA + AD + BD + CD = ABCD + 2*(A+B+C+D)
(sum of all possible subsets of size 2 among 5 winners) = ABCDE + 3*(A+B+C+D+E)

hmmm I would have expected it to be all combinations. For example, in the 4 winner case I would expect the left side to contain terms like ABC

If you define V(A) as 1/number of voters who voted for a and E(AB) as 2 * number of voters who voted for both A and B * V(A) * V(B), then voter who votes for A, B and C has load of V(A)+V(B)+V(C). Square it and you get V(A)^2+V(B)^2+V(C)^2+2V(A)V(B)+2V(B)V(C)+2V(C)V(A). Now group the parts in the first half of the expression across all voters who voted for a certain candidate. If you sum it up you would get voters who voted for A * V(A)^2 + voters who voted for B * V(B)^2 + … = V(A) + V(B) + … Group parts of the second half across all voters who voted for a certain pair of candidates and sum it up and you get voters who voted for both A and B * 2V(A)V(B) + … = E(AB) + …
Therefore, the sum of squared loads with candidates A, B, C is V(A) + V(B) + V(C) + ... + E(AB) + E(BC) + E(AC) + ...

I’ve just done my own calculations which confirm the D’Hondt equivalence of adding the sum of squared loads for the candidate set to the sum of squared loads for each individual candidate.

I think it’s likely to give largely the same results as CFAT, but I presume it will also be subject to monotonicity failures as well, without something like approval removal. However, one of CFAT’s problems I think is that the splitting of voters damages independence of irrelevant ballots, so this method might be superior.

Edit - Regarding non-monotonicity, given the following ballots:

1: A
1: B
1: AB
3: C

Without any approval removal, C would be favoured (i.e. would get more than half the seats). With 30 seats, A and B would get 7 each (totalling 14), whereas C would get 16. This is my calculation if it makes any sense to you, but you can do your own working.

Edit 2 - That example was borrowed from Warren.

Do you have a feel for what the source of the failure of monotonicity is? Not specific examples but theoretically. I have not really studied the Phragmen methods much. Are there any that are monotonic? Monotonicity is a voter based metric and the loads have winners as their source. Is there some sort of fundamental incompatibility here.

My system, Sequentially Spent Score, is almost the exact opposite theory. Phragmen gives each winners a load to distribute on voters. SSS gives each voter score to distribute/spend on winners.

Maybe there is no insight here. Just thinking out loud.

The source of the non-monotonicity is a lack of “balance”. With the example I gave:

1: A
1: B
1: AB
3: C

It would be better for the AB voter to be split giving:

1.5: A
1.5: B
3: C

The proportional result should be for A and B to get a quarter of the seats each, and for C to get half. And in this case every voter would get the same load. The C voters get half the load of the A and B voters per elected candidate but they get twice as many.

But with the original ballots, the AB voter gets twice the load of the A and B voters, which generates an imbalance. If candidates are elected in the proportional 1:1:2 ratio, the sum of the squared loads of the 3 A and B voters would be higher than that of the 3 C voters. Reducing the proportion of A and B candidates that are elected and increasing C (giving a disproportional result) can then actually bring the sum of squared loads down.

CFAT and the modification suggested here by @matijaskala make the method slightly more monotonic because extra approvals count along with balance, but it doesn’t cancel out the non-monotonicity completely.

According to my calculations, (A+B)*(AB-A-B)/(3*A+3*B-AB)+(B+C)*(BC-B-C)/(3*B+3*C-BC)+(C+A)*(CA-C-A)/(3*C+3*A-CA) + 2*(A+B+C) should pass that example. But it’s quite possible that I made an error somewhere.

EDIT: there was another error that made the expression collapse into Sainte-Laguë instead of D’Hondt.

That’s looking quite complicated. And assuming it works, was it designed specifically for that example, or do you think it might work more generally? Was there an intuition behind it that helped you find that particular formula?

It was designed for cases that involve evenly split voter groups.

The intuition was that if
1 A
1 B

has the same value as
2 AB
x A
x B
2-2x AB

should have the same value as well.

I don’t think this is a great starting point.

  1. An election result in which two candidates are unanimously approved is obviously better then one in which voters are in two equally-sized polarized camps of only approving A or only approving B.

  2. If a proportional voting method cannot prefer any election other election result to the one in which half of voters approve of A and half approve of B, then that forces the method to fail either monotonicity, IIB, or both

Well, it does prefer 2 AB once you turn it into D’Hondt.

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It looks like this new method fails ULC. Universally liked candidates seem to help candidates with less approvals.

This passes ULC:

1/(4/(AB-A-B)-1/min(A,B)) + 1/(4/(BC-B-C)-1/min(B,C)) + 1/(4/(CA-C-A)-1/min(C,A)) + A + B + C

It looks pretty horrific, but if it works it works. Is it still monotonic and would there be a four-way tie for the fifth seat here?

Do you know if it passes independence of irrelevant ballots? And is it still D’Hondt for party voting? All these questions I know, but you could be onto something.

Yes, it passes your example and it is D’Hondt. It doesn’t pass IIB but it’s possible that it passes its weaker version.

It’s not fully monotonic and I suspect it just trades some non-monotonicity cases for others. ABC+A+B+C with approval removal seems better than this.

Found this. There are some non-monotonicity cases, but at least it’s relatively simple:

1/|voters who approved A| + 1/|voters who approved B| + |voters who approved both A and B|/(|voters who approved at least one of {A,B}| * min(|voters who approved A|, |voters who approved B|))

1/V(a) + 1/V(b) + V(a and b)/V(a or b) * min(V(a), V(b))

Looking at this, I found an example where this is better than CFAT.

The following is a D’Hondt tie:

2 to elect, approval voting

2 voters: A
1 voter: B

AA and AB tie under ordinary D’Hondt party list, Ebert + CFAT, and Ebert + your modification.

However, D’Hondt and Ebert + CFAT don’t give the equivalent tie where there are four winners.

Although AAAB should win, there should still be a tie between AAAA and AABB. From my calculations, CFAT fails to produce this whereas the modification from @matijaskala does. This might not seem important as it’s not going to affect the winning set, but it wouldn’t be hard to contrive an example where it would be relevant similar to what I did here. I haven’t proved it works in the general case though where you multiply the number of winning candidates by an arbitrary constant from any D’Hondt tie case.