# 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"
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)