A compromise between Vote Unitarity and Thiele PR methods

Methods that pass Vote Unitarity seem to return more representative outcomes than variants of Thiele such as RRV and SPAV. Vote Unitarity methods tend to pass ULC, for example, which Thiele fails. They also don’t use the same candidate to satisfy multiple voting blocs, as PAV does in the following case I described in a different post:

In this case, PAV uses the candidates A, B, and C to satisfy the top 4 Hare quotas of voters. In contrast, Vote Unitarity methods would make the election of A, B, and C cost the top 4 Hare quotas of voters 3 Hare quotas in total, most likely split between the two. Electing F and G would cost the bottom two rows of voters 2 Hare quotas, leaving them with nothing. So the final seat would go to either D or E.

However, as a candidate’s election can be forced with less than a Hare quota, and sometimes much less, Vote Unitarity methods are also more vulnerable to Hylland free riding and vote management than Thiele variants. Here is a case where they can be used to reverse a majority:

Electing two members of party B costs the bottom row of voters 40 votes, leaving them with 11. So all 3 A candidates get elected with less than a full quota.

With Thiele variants such as RRV and SPAV, there is no incentive for a party with no support from non-party voters to divide themselves in such a manner. (I have termed this the “Limited Resistance to Hylland Free Riding” criterion.) This allows partisan voters more freedom to distinguish between candidates in their party, as they don’t just have to follow instructions from the party’s leadership, something that vote management entails. This is largely due to these methods having D’Hondt party list cases, although that isn’t sufficient to guarantee this.

I will now describe a method that attempts to compromise between the two methods. I will only describe the Approval case for now, as I have not yet decided how to extend it to Score.

All ballots start with a weight of 1. Select the largest quota that makes it possible to elect a candidate without going below the quota. The candidate that can be elected with this quota is elected. Call the quota q1. To elect the candidate, each ballot must pay weight*[q/(weighted total votes for the elected candidate)]. Proceed to the next round. The next candidate who is elected is again the one that can be elected with the largest quota size (call it q2) if the voters for the candidate elected in the previous round only had to pay q2 to elect that candidate. In later rounds, again, the candidate who is elected is the one who can be elected with the largest quota size, qn, if voters for each candidate elected in the previous round only had to pay qn to elect their candidate.

As an example, consider a 4 winner election with the following ballots:
20 ABCD
10 AE
q1=30, and A wins, because A can be elected if the quota is 30, and no other candidates can.
B, C, or D can be elected second with a quota of 12, since if A were elected with a quota of 12, then the weight of the ballots would be:
12 ABCD
6 AE
E can only be elected with a quota of 7.5, since if A were elected with a quota of 7.5, then the weight of each ballot would be:
15 ABCD
7.5 AE
So B wins, and q2=12.
C or D can be elected third with a quota of 7.5, since after A was elected with a quota of 7.5, the ballot weights would be as listed above, and after B was elected with a quota of 7.5, the ballot weights would be:
7.5 ABCD
7.5 AE
E can also be elected with a quota of 7.5. It seems clear that if E is elected, C will be next, and if C is elected, E will be next, so the winners are A,B,C,E.

A potential modification to this method would be to set a maximum for the quota in the early rounds. If it is possible to exceed this maximum quota, just elect the score leader for that round if the chosen maximum is used as quota. The final quota is necessarily smaller than a Hare quota, so a Hare quota would probably be a good maximum to choose. If this is done, this method behaves like Sequentially Subtracted Score until it is no longer possible to elect candidates with full Hare quotas. I will probably extend this method to Score in a way that maintains this when such a quota cap is used.

What if you used STAR on all candidates who fall below a Hare Quota of points? That might ensure the majority can win the final of the 5 seats, without unnecessarily polarizing the other seats.
Edit: To perhaps avoid majorities manipulating this to get more seats, just run STAR on the final seat. At worst, the majority gets one more seat. I don’t know if this is enough to ensure majorities always get their fair share of seats if the number of seats to be elected is very large however, though all major PR proposals seem to involve only a few members being elected within a district.

This is the much better way to do it for public marketing, since Sequentially Subtracted Score itself is simple enough to explain to people, and this can be thought of as the complexity behind the system, similar to fractional transfer in STV.

Notably, this method passes ULC while also having a party list case of D’Hondt, which is an unusual combination of properties. In the case I described where PAV uses the same reps to satisfy multiple proportionality guarantees, this method does not do the same thing, however I do not know whether that is always true.

The optimal version of this method is probably one such method that I have frequently thought about but ultimately rejected because I couldn’t get it to pass IIB.

Voters have 1.0 weight that can be distributed among all the candidates they approve of. Find the optimum distribution of voter weights to candidates such that the minimum amount of total weight held by a winning candidate is the greatest.

Is it even possible to get ULC and IIB?

Maybe. Maybe not.

@fsargent can u please get rid of this stupid 20 character rule?

Warrren invented such a system last spring. I am not sure if he tuned it. It came out of my discussions with him about Vote Unirarity. Basically you do single value decomposition on the score matrix to find the best winner.

How well would this system work with multi-member districts?

I think it assumes them

How easily can this optimum version be computed?

With the sequential version, the exact formula for the next quota gets increasingly more complicated as more winners are chosen. However, the quota can easily be approximated. It must be between 0 and Votes/round. If you plug in an estimate and calculate the scores for each candidate at the current round for the estimated quota size, then it is exactly correct if the leading score is equal to the estimate. If the leading score is smaller than the estimate, the estimate is too large. If the leading score is larger than the estimate, the estimate is too small. (Also, the leading score provides another bound for the true quota in the direction opposite from the estimate, but that isn’t always useful.)

Some information here https://en.wikipedia.org/wiki/Singular_value_decomposition#Calculating_the_SVD

This isn’t the same thing as what I was talking about, but OK… Not sure how his method is an optimal version of the sequential method @Marylander came up with in the same way that mine is.

No Warrens SVD system is optimal not sequential. It is also not vote unitarity preserving.

The problem with this idea is that the minority can run one more candidate than the number of seats they seek to win, and crowd out the runoff for the final seat. So in the example:

B3 wins with STAR, fixing things. But if there is an additional candidate A4 that the 16 A3 voters also max-score, then the result would be:

16 votes: A3 5 A4 5
11 votes: B3 5

A3 and A4 go to the runoff, ensuring an A candidate wins this seat.

Although if you ran a modified “Vote Unitarity PR” runoff, you could have the first spot in the runoff go to the highest-scored candidate, then subtract the number of points the voters gave that candidate from their ballots, and then take the next highest-scored candidate for the second spot. This defeats Vote Management. I’ll make a separate topic, sorry.

How would this method fix the issues in the example?

Well, the first winner in that example (converted to Approval ballots) would be B1 (q1=51). The second winner would be B2 (q2=25.5). The third winner would be either A1 or B3 (q3=17), and the fourth would be the other of A1 or B3 (q4=17). The fifth winner would be A2 (q5=16). So A wins 2 seats and B wins 3.

How is this calculated?

From what I can tell, the other way to run this method is that whenever someone wins with less than a Hare Quota before the final seat, all ballots that spent points prior to the non-Hare Quota candidate’s election are restored the number of points they scored all Hare Quota-meeting candidates (or could it just be all candidates?!) the maximum score times the fraction of a Hare Quota that the non-Hare Quota candidate did not meet.
In your example, Party B receives 3 votes back after the first Party A member wins, because they were elected 3 votes short of a H.Q., then 4 from the fraction of the other Party A winner’s unfilled Hare Quota. The combined 11 + 7 = 18 votes lets them win the final seat against Party B’s 16 votes. Would this way be easier to implement for public elections?
We could call it “deficit handling.”

I’m not sure whether it’s possible for a ballot to have more than 1.0 weight because of the restoration here. Actually it ought to be, so I’ll make another post for this as well.

Sorry, I am really not understanding how you rewieght. The selection seems to take the highest reweigted sum of score like most other utilitarian systems.

That is exactly what you would get from RRV. Can you be more explicit in the calculation? Maybe just give the logic.

I think you might be saying that the reweighting depends on the scores of the runners up. Brams’ new method is like this. That implies that it is not immune to clones which essentially brings back vote splitting. Hylland free riding is an issue but not near as important as clone immunity. In this case the cure is worse than the disease. Its like getting PR through partisan voting. Just not worth it.

In that specific example, yes, it behaves like RRV. However, the method is still clone immune. What happens is: for each round, there is a specific value, q, associated with each unelected candidate. The only scores that an unelected candidate’s q value depends on are the scores given to that specific unelected candidate and the scores of the previous winners. A candidate’s q is defined as the unique value such that if the election were restarted with the following rules: (1) the candidates elected in the previous rounds of the real election are elected in the same order in the hypothetical replay, and (2) the surplus handling is the same as in SSS, except with q used as the quota instead of the actual Hare quota, then (3) this unelected candidate will have exactly q points in the round after all previous candidates have been elected. The candidate elected next in the real election is the one with the greatest q.

In the “capped quota” variant, the procedure described above is only followed when q would be smaller than the cap. Otherwise, the procedure to follow is similar, except the quota size to use in the hypothetical election is the cap, and the candidate to elect is the one who would be elected with the greatest surplus. When this cap is set to a Hare quota, the method behaves like SSS until no candidate can meet quota.

The problem with that approach is that when you restore some weight to other ballots, then it raises each candidate’s weighted score, so the candidates will get closer to a full quota. This means that you have given the voters too much weight back, so you must take some weight away again. Also, this could change which unelected candidate has the most weighted points, so declaring a winner before restoring the extra points might be a mistake. You could try taking some weight away from the ballots, and not electing anyone until you know that all winners cost the same, and also that there will be no surplus. And that is basically what the capped version of my method does. However, repeatedly choosing the leading weighted score as your next estimate for the quota is a poor iterative estimation procedure; it’s not even guaranteed to converge.