Talk:Huntington–Hill method

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Who is this method named after? --Palnatoke (talk) 09:32, 26 September 2008 (UTC)[reply]

It was introduced by Hill, & advocated-for by Huntington. I call it "Hill-Huntington", because Hill was its introducer. 2600:6C55:7900:2B8:84F0:30BF:A636:A439 (talk) 09:29, 28 August 2023 (UTC)[reply]

After this recent

.

The constraints applied to this problem are that the total number of Representatives is fixed and determined in advance by law: 435, and that each state, even the least populous, must get at least one Representative.

Let

${\displaystyle K\ }$ be the number of states (currently 50).

${\displaystyle 1\leq k\leq K\ }$ be the index of k^th state. It doesn't matter how they're ordered.

${\displaystyle P_{k}\ }$ be the agreed census population for the k^th state.

${\displaystyle P=\sum _{k=1}^{K}P_{k}\ }$ is the total population of all K states (excluding DC and the territories).

${\displaystyle N_{k}\ }$ is the number of Representatives for the k^th state that we are trying to determine.

${\displaystyle N=\sum _{k=1}^{K}N_{k}\ }$ is the total number of Representatives in the House for all K states (excluding DC and the territories) which is currently 435.

${\displaystyle q>0\ }$ is the nationwide constant of proportionality or quota ratio for proportionately allocating Representatives of a state as a function of its population.

So, if we could actually have fractional numbers of persons as Representatives,

${\displaystyle N_{k}\ =\ q\ P_{k}\quad \quad \forall \ 1\leq k\leq K}$

${\displaystyle \sum _{k=1}^{K}N_{k}\ =\sum _{k=1}^{K}\ q\ P_{k}=q\sum _{k=1}^{K}P_{k}}$

or

${\displaystyle \ N=qP\quad }$

${\displaystyle \ q={\frac {N}{P}}\quad }$

But, of course, we cannot divide Congressional Representatives into fractions even if we might like to tear them apart on occasion. Each states House delegation must be an integer number of people at least as big as one. Wouldn't this mean:

${\displaystyle N_{k}\ =\lceil q\ P_{k}\rceil \quad \quad \forall \ 1\leq k\leq K}$ ?

where

${\displaystyle \lceil x\rceil =\min \,\{n\in \mathbb {Z} \mid n\geq x\}}$ is the

ceiling function

(which means always round up).

Now if q > 0 was arbitrarily small (but positive), then each state would get 1 Representative. It wouldn't be particularly well apportioned and it wouldn't add up to N=435. We want

${\displaystyle N=\sum _{k=1}^{K}N_{k}=\sum _{k=1}^{K}\lceil q\ P_{k}\rceil \ }$

Then couldn't q be increased monotonically, thus increasing some of the N_k and the total ${\displaystyle \sum _{k=1}^{K}N_{k}=\sum _{k=1}^{K}\lceil q\ P_{k}\rceil \ }$ until it reaches the legislated N=435 value? Would that not be the meaning of proportional representation with the constraints that N_k must be an integer at least as large as 1? How is there any paradox in this method (assuming that, as q increases we don't have two states simultaneously increasing their integer N_k and the total jumping from 434 to 436) and where the heck does that

come from? How does that possibly have anything to do with true proportional allocation of a fixed number of seats in the House?

I can tell that this q is is close to !/A in the article, but I have no idea where this geometric mean of N_k and N_k+1 comes from. What motivated Huntington or Hill or whomever to use that mathematical construct? What does it have to do with accurate proportional allocation of a finite resource (seats in the House of Representatives) among the population?

Can someone explain this? 96.237.148.44 (talk) 18:35, 14 February 2009 (UTC)[reply]

What you describe is Adams' method (no article yet, but mentioned in Apportionment (politics)). Huntington/Hill can be formulated the same way, with the ceiling function replaced by geometric rounding:

${\displaystyle N_{k}=\left\downarrow qP_{k}\right\uparrow \quad \forall \ 1\leq k\leq K}$

where

${\displaystyle \left\downarrow x\right\uparrow ={\begin{cases}\left\lfloor x\right\rfloor &\mathrm {if} \ x<{\sqrt {\left\lfloor x\right\rfloor \left(\left\lfloor x\right\rfloor +1\right)}},\\\left\lceil x\right\rceil &\mathrm {else.} \end{cases}}}$

This is more proportional than Adams' method because it doesn't favour small states that much, but still guarantees N_k to be an integer at least as large as 1. The most proportional

divisor method is the Sainte-Laguë method

(or Webster's; round to nearest rounding function; used for US congressional apportionment in the past), but it needs additional provisions to guarantee at least 1 seat for every state.

Huntington/Hill minimizes the relative deviations of seats per population, while Sainte-Laguë minimizes the absolute deviations of seats per population. --84.151.17.20 (talk) 22:52, 17 November 2009 (UTC)[reply]

The Hill-Huntington allocation rule was designed to maximize proportionality, & it does (reckoning s/v variation by ratio rather than difference).

Hill-Huntington gives an allocation such that if a seat is transferred between any two states, that will never bring their s/v ratios closer to eachother. In fact it will move them apart.

i.e. It's the most proportional allocation, when s/v variation between the 2 states is measured by ratio rather than difference.

Webster, which was used over two intervals, one starting in the early 19th century, & the other ending in 1940, accomplishes the same thing, except that it gives an allocation such that any transfer of a seat between any two states will never bring their s/v ratios closer together (where that distance is reckoned as difference rather than rati

I have to admit that measuring distance between 2 states' (or parties') s/v values by ratio makes more sense to me, seems more right.

But Webster has advantages that outweigh that aesthetic consideration:

1) It's a lot simpler.

2) It's only half as biased as Hill-Huntington.

...& which is more important?: The ratio vs difference aesthetic consideration, or less bias, less systematic favoring of larger or smaller parties? I say unbias is much more important.

Suggestion: Change back to Webster.

Webster is very slightly biased in favor of larger states.

Hill-Huntington is about twice as biased, in favor of smaller states.

Webster is the most unbiased apportionment-method that has ever been used or proposed.

...or was until about 17 years ago, when I introduced Bias-Free.

Bias-Free has no bias.

...but it's more complicated than Hill-Huntington, & its derivation wouldn't be as easy to explain as that of Webster.

But, on the other hand, if someone can accept Hill-Huntington being more complicated than Webster, then might they not, as well, accept Bias-Free being more complicated than Hill-Huntington?

For Bias-Free, substitute (1/e)((b^b)/(a^a)), for sqrt(ab).

...where a & b are the two seat-numbers between which the rule is being applied.

...& e is the base of the natural-logarithms.

If simplicity were important, Webster wouldn't have been replaced with Hill-Huntington.

So then, replace Hill-Huntington with Bias-Free.

There was at least one academic journal-article about Bias-Free. I was notified about one. At least the notification was about an apportionment method that was named after me & another person. ...& Bias-Free is the only apportionment method that I've introduced. I didn't get a chance to look at the article, because I couldn't afford the price of a year's membership with the article-providing service. I don't remember the other person's name, or the name of the journal or the title of the article (if they were given in the notification).

Of course, until I find that article, I'm not in much of a position to propose Bias-Free. 2600:6C55:7900:2B8:BD0B:124D:7AF1:4203 (talk) 16:42, 28 August 2023 (UTC)[reply]

Is this different than the actual apportionment? If it is just an example, the use of a fictional situation would be better. If this is an example AND for comparison, then let's have the numbers we are comparing against. Jd2718 (talk) 15:25, 22 April 2011 (UTC)[reply]

Actual apportionment was the same except: Deleware and Vermont had 1 and 2 reps, respectively, instead of the 2 and 3 this method would have granted them, and Pennsylvania and Virginia had 13 and 19 respectively, instead of the 12 and 18 they would have had under this apportionment system. Still, this is not an article about the 1790s... perhaps a fictitious situation would be better than this near-

Original Research Jd2718 (talk) 15:35, 22 April 2011 (UTC)[reply

]

Israeli Knesset example mathematics

Refactored

(?) from article

2601:602:C800:36D6:9957:CEDB:EB20:B88D (talk) wrote:

Please note: the below Israeli Knesset example is not mathematically correct. The Huntington-Hill seats column totals to 119, not 120. 

The correct allocation for the D'Hondt method is:  30    24    14    11    9    8    7    6    6    5.  

Regardless of the actual Israeli Knesset allocation (which may be some modified version of the D'Hondt method?), please check your math and/or use an example which cites an English language reference source.

I moved this here so that we could discuss the concerns.

Firstly, you are right, I mistyped when I transposed numbers from the previous version of that table. United Torah Judaism should have had 6 seats, which would bring the total up to 120 for the Huntington-Hill seats column.

Secondly, while I agree that an english language source for the vote totals would be ideal, I am unaware of any reason why that would be required. I see no reason to seek one out. Mayhaps someone interested could find one?

Thirdly, you are also right that this is not a standard D'Hondt allocation. The person who originally wrote that probably should have clarified that it is actually a modified D'Hondt called the Bader-Ofer System, which allows for spare vote agreements between parties. I will add a note.

Brvhelios (talk) 07:22, 4 May 2021 (UTC)[reply]