# Talk:Schulze STV

## Party list case

Schulze STV's party list case is not D'Hondt because Schulze STV passes the Droop proportionality criterion, which is quota-based. No divisor methods pass quota criteria, so Schulze STV's party list case can't be any of them. Most likely, it is LR-Droop, but I haven't proven that. Kristomun (talk) 10:41, 18 March 2020 (UTC)

- I'm not sure I understand, though I'll defer on this. DHondt guarantees every party will get at least as many seats as it has HB quotas, which also guarantees Droop proportionality, no?

- To give an example, if you had a 2-seat election with 50 votes for Party A and 10 votes for every party from B through Z, then both Schulze STV and divisor methods give A both seats, whereas any largest remainders method will give A only one seat, since that is what the quota rule mandates. BetterVotingAdvocacy (talk) 16:51, 18 March 2020 (UTC)

- User:Kristomun, I'd like to convince you on this point. See W:Hagenbach-Bischoff system, which says:

- "The Hagenbach-Bischoff system is a variant of the D'Hondt method, used for allocating seats in party-list proportional representation. It usually uses the
**Hagenbach-Bischoff quota**for allocating seats, and for any seats remaining the D'Hondt method is then applied so that the first and subsequent divisors (number of seats won plus 1) for each party list's vote total includes the number of seats that have been allocated by the quota.**The system gives results identical to the D'Hondt method**"

- "The Hagenbach-Bischoff system is a variant of the D'Hondt method, used for allocating seats in party-list proportional representation. It usually uses the
- In other words, D'Hondt guarantees every party wins at least as many seats as it has HB quotas. The [Droop proportionality criterion], in the party list case, only says that every party must win at least as many seats as it has more voters than that number of HB quotas. I think you're confusing the quota rule with Droop proportionality here. BetterVotingAdvocacy (talk) 02:54, 21 March 2020 (UTC)

- It should be pretty easy to determine this. Find an election where LR-Droop and D'Hondt disagree. Turn it into a Schulze STV election and see what the outcome is. Either you get the LR-Droop result, or you get the D'Hondt result; either way will strengthen the case considerably. It's something I'd like to do, eventually, but I haven't got around to do yet. Kristomun (talk) 10:36, 21 March 2020 (UTC)

- Perhaps use the "every divisor method fails quota" election in https://rangevoting.org/Apportion.html Schulze STV source code here, if you feel like doing it: https://is.gd/cQpgQG Come to think of it, linking to the source code in the Schulze STV article itself might be a good thing to do. Kristomun (talk) 10:46, 21 March 2020 (UTC)

- That's a good example, since it shows Schulze STV agreeing with D'Hondt, but not Webster or LR-Droop. So, the results in each method for that example: D'Hondt gives 15 seats to D, Webster 12, and LR-Droop 14. All three of them spread the remaining of 18 seats equally to 3 other parties, resulting in a tie in LR-Droop. I think WLOG suppose Party A gets the extra seat from the tie in LR-Droop. Now, for the D'Hondt vs. LR-Droop matchup, it's a difference of exactly one (give D one more seat or A one more seat), so there's a direct comparison; with the D'Hondt>LR-Droop part first, we see 105 voters splitting among 15 candidates, so the minimum held by anyone is 7 votes. With the LR-Droop>D'Hondt part, we see 13 voters splitting among 2 candidates, so the minimum held by anyone is 6.5 votes. So D'Hondt wins the matchup. In general, I think we can see that when you propose giving a party less than its D'Hondt share, it can always divide more votes among its candidates than can its competitors; in fact, this seems to be exactly what it is designed for. Now, regarding D'Hondt>Webster, we need to construct the beatpath. We already saw that 15 seats for D beats 14 seats, so it's obvious that 14 beats 13 beats 12, since that would require every smaller party to get 2 seats. Since they're all 13 votes large, they can do at best 13/2=6.5, while Party D can do 105/15=7 and thus when you propose it gets only, say, 13 seats, it can do 105/13=8.07 votes, etc. So, I think I've given you a strong intuition for why Schulze STV decays into D'Hondt: D'Hondt itself is modeled off of the HB quota in its formula, and the HB quota was invented because it allows larger groups of voters overwhelm smaller groups, similar to vote management. BetterVotingAdvocacy (talk) 16:18, 21 March 2020 (UTC)

- I still don't entirely follow. I'll have to run a D'Hondt vs other scenario through the Schulze STV software someday. The example on rangevoting doesn't work because there are too many candidates and the software just bombs. Oh well. The burden of proof is now on me to show that the conclusion is wrong (if it is), so I'll look more into it when I have the time. Kristomun (talk) 11:28, 25 March 2020 (UTC)

- Keep in mind that in the party list case, you can simplify the calculations so much that it can be done by hand (or even mentally, if the example is really simple). You can think of each pairwise comparison as "should this party get one more seat or that party?" (this directly follows from the fact that each comparison is between winner sets that differ by exactly one candidate). Further, the winner of a comparison is simply determined by checking which of the two benefitting parties has more votes per the max seats they're trying to take in that comparison, because none of the other parties has any preference between those two, and because each of the two parties will prefer to vote manage in favor of the winner set when their party would get more seats. Because this is really equivalent to finding a highest average, we want to think about three cases: when a winner set gives a party their D'Hondt share, and when it goes over or under. If it goes over for one party and under for the other, then the under-party can get a better average i.e. If for two parties A and B, A's proper share is 10 and B's 8, and we compare winner sets with A at 15 and B at 7, we know that based on D'Hondt, A simply couldn't have had that many votes per seat. Maybe let me try to show you why the D'Hondt formula is equivalent to vote management: when a party gets its first seat, its votes are divided by 2. This never made sense to me until I thought of it as the party essentially seeing whether or not it could split more of its votes between 2 seats such that it had more votes per seat than any other party. When it gets that second seat, its votes get divided by 3, because it's looking to play for 3 seats now. In a system like regular STV, when parties have to decide whether to vote manage, they make exactly this type of calculation i.e. can we put more votes in each seat than anyone else. It's just that in D'Hondt, you first get the seats that you can get and then if your attempt to take an additional one fails, no harm, whereas in STV, if you spread too thin you get nothing. (Just to clarify, by vote management in STV I'm referring to the Hylland type where a party might tell half of its voters to bullet vote one of its candidates and the other half to bullet vote the other guy.) I can maybe create some graphics later and link them to you if I can think of how to demonstrate it. BetterVotingAdvocacy (talk) 14:24, 25 March 2020 (UTC)

## Approval case

This is an open question: what Approval PR method does Schulze STV become equivalent to when all voters Approval-style (they rank all candidates either 1st or last)? BetterVotingAdvocacy (talk) 08:30, 1 April 2020 (UTC)