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First, I can't tell you how happy that user:Ruhrjung added this page, and the sainte-lague one, and that we've had all these people working on these pages. I'm going to suggest that we replace the example with the following made-up one: apportioning seats to states for an 8-seat council of New England states. The reasons I'd want to make the change are: 1) the "preferences" (in this case, population) aren't made up, so nobody has to account for "is this a realistic spread of preferences or is this tailored", and 2) it emphasises the difference between d'H and S-L (that's why I chose not to use modified S-L).
However, I'm not going to be bold because Ruhrjung clearly put in a lot of work to make the existing example (whose table structure I have copied) and I don't want to offend him. Is this okay? Which example do you think is better?
d'Hondt method | Sainte-Laguë method | ||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
States | MA | CT | ME | NH | RI | VT | MA | CT | ME | NH | RI | VT | |
Pop. (1000s) | 6,349 | 3,405 | 1,274 | 1,235 | 1,048 | 608 | 6,349 | 3,405 | 1,274 | 1,235 | 1,048 | 608 | |
mandate | quotient | ||||||||||||
1 | 6,349 | 3,405 | 1,274 | 1,235 | 1,048 | 608 | 6,349 | 3,405 | 1,274 | 1,235 | 1,048 | 608 | |
2 | 3,175 | 1,703 | 637 | 2,116 | 1,135 | 425 | 412 | ||||||
3 | 2,116 | 1,135 | 1,270 | 681 | |||||||||
4 | 1,587 | 907 | |||||||||||
5 | 1,270 | ||||||||||||
seat | seat allocation | ||||||||||||
1 | 6,349 | 6,349 | |||||||||||
2 | 3,405 | 3,405 | |||||||||||
3 | 3,175 | 2,116 | |||||||||||
4 | 2,116 | 1,274 | |||||||||||
5 | 1,703 | 1,270 | |||||||||||
6 | 1,587 | 1,235 | |||||||||||
7 | 1,274 | 1,135 | |||||||||||
8 | 1,270 | 1,048 |
Also, whichever example we use, I want to use and explain a table like this:
d'Hondt
method |
Sainte-Laguë method | ||||||||||||
States | MA | CT | ME | NH | RI | VT | MA | CT | ME | NH | RI | VT | |
Seats | 5 | 2 | 1 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 1 | 0 | |
Ratio | 3.65 | 1.96 | 0.73 | 0.71 | 0.60 | 0.35 | 3.65 | 1.96 | 0.73 | 0.71 | 0.60 | 0.35 | |
Diff. | +1.35 | +0.04 | +0.27 | -0.71 | -0.60 | -0.35 | -0.65 | +0.04 | +0.27 | +0.29 | +0.40 | -0.35 |
Go along. I'm busy at work for the next three days - probably without any time to "recover" in front of the computer.
Your proposed addition is quite in line with my thoughts on what's relevant and interesting. I had a thought with the made up number of votes, namely that they make percentage-comparisons easy for the untrained, as the total number of casted votes was set to 100.000. And as quite a few elections in the last 30 years (i.e. in my lifetime:-) has been done with these methods, it never occured to me that one of them ought to be taken as an example. By making up the numbers, the which-and-why question was avoided. :-) My idea was to add rows to the tables with approximately the following information:
const. result in percentage size d'Hondt Sainte-Laguë party 1 party 2, party 3... party 1, party 2, party 3... 1 100% (1) 100% (1) 2 100% (2) 100% (2) 3 67% (2) 33% (1) 67% (2) 33% (1) 4 50% (2) 25% (1) 25% (1) 50% (2) 25% (1) 25% (1) 5 6 7 8 9 10
I had further had the idea to have two sections in the article with two headings: One for comparing the unmodified Sainte-Laguë method, and one (as I had started) for the method with the first divisor set equal to 1.4 - and how unlikely it might now ever seem, I had actually finished that work when my computer locked, and the work was lost, ...and I soured.
Do as you like with the figures. I wouldn't advice you to use any New England example, as the whole wikipedia project already as it is is pretty much US-centered, which isn't always a good thing, and as it might seem odd to list half-a-dozen of non-US countries where the method is in uncontroversial use, and then give an example from USA. However, I'm glad someone more than me has found it relevant to work on these articles, and I'm sure the end result will become pretty good whatever you choose to do of your New England idea.
best regards!
--
Ruhrjung 17:52 28 Jul 2003 (UTC)
Uf! I'm sorry to hear about your computer lockup! I look forward to the text being regenerated. So, I agree about the US-centrism, and I'll drop the New England example, but I still think it's better to use a "real-life" made-up example (EU? AU?) than an out-of-the-blue made-up example.
By the way, do you know about Wikipedia:WikiProject Voting Systems? You don't need to know anything about the project in order to participate, but if we want project-wide standards, then that could be a place we work them out.
See you,
DanKeshet 04:35, 29 Jul 2003 (UTC)
One more step taken, on the outlined road, but the weather is nice, and the summer short up here in the North, why I refuse to hide indoors more. :))
--
Ruhrjung 09:05, 5 Aug 2003 (UTC)
I saw that. The colour is nice. Your attempt to explain the method as if it was largest remainder method is maybe not quite simple to understand, that's true, but if so, it can surely be mended in due time. I'm for instance pretty fond of the following sparse wording, quoted from http://www.barnsdle.demon.co.uk/vote/appor.html:
-- Ruhrjung 13:58, 7 Aug 2003 (UTC)
Would it be useful to mention minor methods like Hill's method? (Hill's method is used for assigning representatives to US states, and has the property that every "party" gets at least one representative.) Rob Speer 08:04, Jul 17, 2004 (UTC)
Im having trouble understanding the exact rationale behind these methods. Why do seemingly provisionaly choosen divisors result in a proportional result? Also, why are such complicated ways of allocation used at all; it would seem that simply dividing the number of votes each party won with the total votes and multiplying this with the number of available seets, rounded down, and then largest remaining fractions up to the remaining number of seets chosen (if you write this out, it is mathematically equivalent to Hare quota) would by definition result in the most proportional results, and its the most intuitive - mathematically most evident - way of doing this. -- 195.29.116.30 04:54, 19 July 2006 (UTC)
____________
Let’s suppose those figures:
A: 340,000 B: 280,000 C: 160,000 D: 60,000 E: 15,000
The result using Sainte-Laguë method would be: 3, 2, 1, 1
If we use D'Hond method, then: 2, 2, 2, 1.
It seems to contradict the idea of a 'more proportional' system.
Thanks. —Preceding unsigned comment added by 81.35.196.9 ( talk) 15:03, 13 November 2008 (UTC)
The table calculates modified S-L with 1.4, 3.4, 5.4,.... This results in a top row with a divisor of 1.4, which is incomparable with all the other tables. Why not use the mathematically equivalent series 1, 2 3/7 (2.429..), 3 6/7,...? This would lead to more comparable numbers (although it would take a footnote to explain it.) 187.143.7.74 ( talk) 14:49, 12 February 2010 (UTC)
I have radically changed the part which previously said: "D'Hondt favors the merging of parties, while Sainte-Laguë favours neither merging nor splitting parties which expect to gain more than 1 or 2 seats. (It does favor splitting of very small parties – expecting to gain only 1-2 seats – into still smaller ones). Modified Saint-Laguë prevents this splitting advantage for small parties, while remaining impartial towards party size for all larger parties" on the grounds that it is not true. Just to illustrate from the current example, if the Yellow party split its 47,000 votes into 6 new parties with about 7,800 votes each, it would win 6 seats in both unmodified Sainte-Laguë and modified Sainte-Laguë (with unmodified Sainte-Laguë it might even successfully do a split into 7 parties with 6,700 votes each so going from 4 to 7 seats). The reality is that Sainte-Laguë often rewards splits, even for large parties, rather like the Hare quotas in Hong Kong. -- Rumping ( talk) 16:45, 22 May 2015 (UTC)
For missing references and other useful material (including mathematical properties of the apportionments achieved and other algorithms always yielding the same apportionment) see http://de.wikipedia.org/wiki/Sitzzuteilungsverfahren . -- Wegner8 07:13, 19 October 2017 (UTC) — Preceding unsigned comment added by Wegner8 ( talk • contribs)
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Hi
not sure on this but I'm not sure the huntington hill example is correct.
because the divisor is n(n+1) the first number in the series is infinity - hence all who meet the threshold get at least one seat. so the first round should both eliminate two of the parties, and assign four seats to the other parties leaving six to be allocated. these will go four to yellow, one to white, one to red, giving overall allocation of 5,2,2,1. In this example it is the same seat allocation as d'hondt and modified S-L.
-- Dirtyrottenscoundrel ( talk) 09:32, 23 October 2020 (UTC) Dirtyrottenscoundrel ( talk) 09:32, 23 October 2020 (UTC)
@ Closed Limelike Curves you are removing the Imperiali quota (post(k) = k + 2) from the divisor table (and elsewhere) without explanation in your edit comments. Please explain that here. Thank you — Quantling ( talk | contribs) 20:52, 26 January 2024 (UTC)
@ Closed Limelike Curves you've renamed several of the methods discussed so that the wikilinks take us to pages that are named differently. As evidenced by the choice of the names of these linked articles, the new names appear to be inferior to the names that they have replaced. But likely you have a reason for preferring the names that you are using ... what is it? Thank you — Quantling ( talk | contribs) 21:11, 26 January 2024 (UTC)
@ Closed Limelike Curves et al: The new column in the table "Allows zeros" might need some re-thinking. Any of the methods that gives post(k=0) = 0 can be said to not allow zeros in that every alternative will get one representative before any alternative gets a second alternative. However, if giving one representative to every alternative makes too many representatives then a floor number of votes is established and only those alternatives with that many votes will get a representative. In this case, some alternatives will get zero representatives even though "Allows zeros" is listed as "no". — Quantling ( talk | contribs) 21:26, 26 January 2024 (UTC)