[EM] a question about apportionment

Kristofer Munsterhjelm km_elmet at lavabit.com
Sat Apr 16 00:10:34 PDT 2011

robert bristow-johnson wrote:
> On Apr 15, 2011, at 8:59 PM, ⸘Ŭalabio‽ wrote:
>>     “Owen Dalby” <Owen.Dalby at Gmail.Com>:
>>>     I apologize if I am asking a dumb question, but would appreciate 
>>> any honest and practical advice from this list. I am conducting an 
>>> election among a group of colleagues who are all graduates of a 
>>> fellowship program. 45 people will vote on perhaps 30 candidates for 
>>> roughly 15 seats.
>>     If these people are not paid, thus it costs group nothing for 
>> their services, just let everyone wanting to have a seat.  If money is 
>> an issue, one should think about the ideal size of the legislature:
>>     If the legislature has 1 tier, then the size of it should be the 
>> squareroot of the electorate.  The squreroot of 45 is:
>>     7
> how well does this work for a large population?  if that rule applied to 
> the U.S., we would have about 17,500 in the House of Representatives in 
> Congress.  might be a little unwieldy.
> would there be some threshold where this rule changes to one where the 
> legislature is a smaller portion?  i s'pose, if i think about it, i can 
> think up an asymptotic function that approximates sqrt(N) for small N 
> and k*N for large N where k is the constant of proportionality.

There are two competing issues here. First, you should have enough 
people to fairly represent the population. I'm not sure how sqrt(n) 
comes into play, but you can consider the absolute lower bound in this 
manner: if you have only one district, then in the best case, you'll 
deny 1/(numseats + 1) of the people representation. If you don't obey 
quota, the fraction may be larger.

So, for instance, with 435 seats and 100 million voters (I'm not sure 
what the US turnout is), there is no way short of weighted voting that 
you can not leave 230 000 unrepresented, if the right 230k happened to 
be a bloc of its own. In reality, Plurality is a lot worse, of course.

Second, you should have a legislature that is small enough that things 
actually get done, and I imagine this is why the US House has been 
frozen at 435. At some point, the other members of parliament or 
Congress stop being people you know (if you're an MP or rep) and start 
becoming just "other people". If you're going to believe monkeysphere 
extrapolation, that number is about 150. As you go further and further, 
the effect increases. I imagine the problem will have greater 
consequences if the parliament is consensus-based than if it's 
party-based or majoritarian. Party loyalty in general could work as a 
sort of "chunking" (I know these people, and then there are [other 
party] whose individual members I don't need to care about), as could 
strict discipline ("it doesn't matter whether I know the other reps 
because I have to vote the way the Central Committee wants anyway").

Note, though, that there are some parliaments and legislatures that are 
even larger than the House. The UK parliament has 650 members, and the 
Chinese has 3000, though the latter is hardly a democracy.

> this is interesting.  in the U.S., the number of Representatives is 435, 
> no matter what the population is, but they have to be allocated 
> proportionally in some sense of the word.  this is called 
> "apportionment" and is done to decide how many Representatives go to 
> each state.  once that is done, states that are apportioned more than 
> one representative have to go through a another slugfest (which is more 
> local and more subjective) in how that state is Redistricted.

Which of course in most states leads to the beast of Gerrymandering. I'd 
say dissolve the problem by having multimember districts (it worked for 
New York's local elections up until the second red scare), but it may be 
very hard to actually get through. People in power rarely want to give 
up their power, after all.

> anyway, the rule they currently use is this Huntington-Hill method.  and 
> i've been wondering exactly how the mathematics work in deriving it.  as 
> anonymous IP, i posted on 14 February 2009 (UTC) a 
> reasonably clear mathematical posing of the same question to:
>     http://en.wikipedia.org/wiki/Talk:Huntington%E2%80%93Hill_method
> using Wikipedia's LaTeX markup.  so it's clearer there than here.  all's 
> i can say is that the conceptually simplest method of proportional 
> allocation, with a fixed number of Representatives and that guarantees 
> each state a minimum of one Representative is:
>    Number of Reps for State k:  N(k) = ceil( q*P(k) )
> where
>    SUM{ N(k) } = N
>     k
> and
>    SUM{ P(k) } = P
>     k
> and where N is the total number of representatives (now 435 in the U.S.) 
> and P is the total population of the electorate (or of those 
> represented, in this case we're counting non-voters) and P(k) is the 
> population of each state.
> q starts out as arbitrarily low (so the SUM{N(k)} is less than N) and is 
> monotonically increased until SUM{N(k)} is equal to N.
> now, this just seems to me to be the most consistent rule that is 
> applied to every state that makes it proportional (to an integer number 
> of representatives for each state) and makes certain each state gets at 
> least one representative.

Don't use ceil(...). That makes the method into Adams's method, which 
has a significant bias for small states (or parties, when it's used for 
party list). Instead, you should use round(), which makes the method 
you've stated into Webster's method.

The immediate problem with this is that some states now get zero seats, 
so you have to give every state at least one seat, *then* you adjust the 
factor q. To do so will introduce a small-state bias, but it's 
unavoidable. So the method would be something like:

  Number of reps for state k: N(k) = 1 + round(q * P(k)),

then if there are s states and n seats, set q so that
  ( SUM (k = 1...s) N(k) ) = n.

The House actually used Webster's method (from 1842 to 1852 and from 
1901 to 1941). I still think it's better than Huntington-Hill (as does 
Peyton Young, http://www.rangevoting.org/pb88.html ).

> can someone explain to me how the Huntington-Hill method is better.  in 
> this section:
> http://en.wikipedia.org/wiki/United_States_congressional_apportionment#The_Method_of_Equal_Proportions 
> they say it's because "the method ... minimizes the percentage 
> differences in the size of the congressional districts."  can someone 
> point me to a proof of that?  i asked the same question in that 
> article's Talk page and have been unsatisfied with the responses.
> can anyone here (like Warren) spell this out?  what so good about 
> sqrt(n*(n+1)) instead of just n?

It depends on the metric you use for disproportionality. Say you have a 
desideratum, and as usual, the number of seats for state k is N(k) and 
the population is P(k), and the desideratum is:

  "If the difference (error measure) between f(N(k), P(k)) and f(N(r), 
P(r)) for some pair of states k and r can be reduced by moving a member 
from r to k or k to r, then that should be done,"

i.e. that the method should at least be locally optimal with regards to 
a certain error measure, in that it's impossible to reduce that error 
measure by moving seats, then

  - Huntington-Hill satisfies this where f(N(k), P(k)) is P(k)/N(k) and 
the error measure is the greater of f(N(k), P(k)) and f(N(r), P(r)) 
divided by the smaller (relative error),

  - Webster satisfies this where f(N(k), P(k)) is N(k)/P(k) and the 
error measure is the absolute difference, i.e. |f(N(k), P(k)) - f(N(r), 

  - A method (Dean?) based on the harmonic mean satisfies this where 
f(N(k), P(k)) is P(k)/N(k) and the error measure is the absolute difference.

The proof goes generally like this: First the proof shows that the 
method in question can be expressed as a party list algorithm. That kind 
of algorithm takes the greatest population remaining, adjusts it by 
division, gives the state a seat, and then repeats until done. See 
http://en.wikipedia.org/wiki/Sainte-Lagu%C3%AB_method for the version 
equivalent to Webster. Second, it shows that if you assume there is a 
pair of states where you could improve the pair by moving a member from 
r to n, then n would have been higher in the queue given by the repeated 
divisions and thus would have got the seat anyway.

For more details, see Huntington's proofs in particular (proofs around 
p. 8, marked as 90): http://www.rangevoting.org/Hunti28.pdf and Warren's 
page on apportionments in general: http://rangevoting.org/Apportion.html .

Finally, note that all the divisor methods may violate quota, but that 
such violations are rare for Webster (and I think Huntington-Hill). All 
divisor methods can in certain cases violate quota, but no method that 
obeys quota can be monotone in the sense that if people move from state 
A to B, it is never the case that B gains representatives at the cost of A.
Also, when I speak of Webster above, in the proof section, I mean the 
one that does not enforce the 1-seat minimum. Warren calls that one 
"Webster0" and the one that *does*, "Webster1" on his simulation page, 
http://www.rangevoting.org/BishopSim.html .

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