# Possible improvement to PAMSAC (gets rid of CFAT)

The Coin-Flip Approval Transformation (CFAT) was added to PAMSAC to improve monotonicity and specifically fix the following example:

2 to elect

x voters: ABC
x voters: ABD
1 voters: C
1 voters: D

Under Ebert’s Method CD would be elected for any value of x. By applying CFAT, AB would be elected for high enough x, and it also changes it from Sainte-Laguë equivalent to D’Hondt equivalent for party voting. However, there is another (probably better) way of doing this. Instead of having to split voters into parts as CFAT does, the weight of a voter is increased by an amount that depends on the number of candidates that they have approved in the set.

If a voter has approved c candidates in a set, that voter’s weight when that set is being considered becomes 2c/(c+1). This, like CFAT, has the effect of turning Ebert’s Method from Sainte-Laguë into D’Hondt equivalent (I can put a proof in another post but will leave it for now). And in the example above AB would be elected for x above a certain amount because there would effectively be more voters when AB is elected.

It would still not be monotonic as things stand. For example:

2 to elect

1 voter: AB
1 voter: AC

Ebert’s Method with this modification would still elect BC and would therefore require approval removal to elect AB (or AC). However, this is where it has an advantage over CFAT. To elect AB, ordinarily you’d have to remove the A approvals from the AB voters. This would make AB and AC equivalent, and AB would win on a tie-break of most approvals. But then it would be impossible for PAMSAC to award the win to AB for any x in the following case even with approval removal:

x voters: AB
x voters: AC
1 voter: B
1 voter: C

But under this new transformation, while approvals would still be removed as before, the weight of a voter can still be based on the number of approvals pre-removal. So:

1 voter: AB
1 voter AC

When considering the AB result, we’d have:

4/3 voters: AB
1 voter: A

And after approval removal this can become:

7/6 voters: A
7/6 voters: B

This beats BC without the need for a tie-break, making the victory far less fragile and it isn’t as susceptible to the single B and single C voters as before. This means that the method would now be strictly monotonic (every approval for a set of candidates would improve it’s Ebert score) than than weakly (where an approval can’t harm it).

Maybe I am misunderstanding but if it is dependent on the number of approved candidates does it not fail Independence of clone alternatives?

It’s dependent on the number of approved candidates in a potential winner set, so it would not fail. If I approve candidates A and B, then for a set that includes just A or just B, my weight would be 1, but if the set included both it would be 4/3.

Anyway, there might be some problems with some aspects of this, so I might need to revisit it.

The new system seems to fail here:

5 to elect

1 voter: U1, U2, A1
1 voter: U1, U2, B1
1 voter: C1, C2
1 voter: D1, D2

There should be a 4-way tie for the final seat. There is with CFAT but there is a tiny difference with this multiplier method.

For U1, U2, A1, C1, D1 you get 280/51 = 5.49. For U1, U2, C1, C2, D1 you get 11/2 = 5.5.

With CFAT Both U1, U2, A1, C1, D1 and U1, U2, C1, C2, D1 give 11. It’s weird and I don’t really get why there should be a difference between the two methods but there you go.

Also the thing about still using approvals a voter multipliers when they have been removed doesn’t seem to work either. For example:

2 voters: A, B
1 voter: C
1 voter: D

Voters approving both A and B would count as 4/3 voters (for sets where both are elected) and approvals could be removed giving:

4/3 voters: A
4/3 voters: B
1 voter: C
1 voter: D

This could give an advantage to the A and B voters, although it’s not necessarily clear to me what in situations such a removal would lead to the lowest sum of squared loads and where this advantage would happen.

What if you replace sum of squared loads if all winners were elected with sum of squared loads if all winners were elected + sum of squared loads if only first winner was elected + sum of squared loads if only second winner was elected + sum of squared loads if only third winner was elected + etc?

This system would collapse into D’Hondt just like your modification.

So if ABC are elected it’s sum of squared loads for ABC + A + B + C? Or do you need to include AB, AC and BC as well?

It’s ABC + A + B + C.

I don’t know whether including AB, AC and BC would produce a proportional result at all. (It might, but I haven’t checked)

Also, assume that the sum of simple loads is W rather than 1. Otherwise it won’t work correctly.

This is interesting because it adds a way to distinguish between the source of the total load. One of my concerns with Ebert’s Method is that it does not distinguish between the source of the load to each voter from each winner. This means that you get the same value for the case when all winners are universally liked as when groups and perfectly block voting quotas.

Max Phragmen breaks this symmetry by taking the max load instead of the square of the sum. The method proposed with ABC + A + B + C would also break this symmetry and likely produce a better result. I would caution that ABC + A + B + C is not really derived. The fact that Eberts method ends up with the squared load is that it simplified from the standard deviation or the mean squared error to W/V. This had to be pointed out to me by @Toby_Pereira. It would likely be a good idea to go back to that step and do them math.

I think the starting quality metric would be something like

(W/V - ABC)^2 + (1/V - A)^2 + (1/V - B)^2 + (1/V - C)^2

but maybe you want the square roots and use the actual standard deviation. This would be different. Maybe I am being pedantic by insisting to go back to first principles but I not sure it is really much math to consider this.

Actually, I derived it from the system I suggested one year ago.

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Awesome. Which one is that?

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It takes some algebra to prove it, but sequential use of ABC + A + B + C is equivalent to that system.

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.