Why Huntington-Hill?

post by jefftk (jkaufman) · 2020-10-25T00:50:33.884Z · LW · GW · 1 comments

The US House of Representatives uses the Huntington-Hill method to figure out how many representatives each state should have. First you give each state one representative, because even the smallest state is guaranteed one, and then you assign the remaining spots, one at a time, to whichever state has the highest "priority". Priority being ratio of its population to the geometric mean of the number of seats it currently holds and the number it would hold if it received this extra seat:

Ps = state population
Rs = state reps

      Ps
---------------
sqrt(Rs*(Rs+1))

Where does this come from? I had a shot at deriving it, and it actually makes a lot of sense. First, we restate the problem has one of error minimization. At every stage, we want to assign the next seat wherever it would most minimize representational inaccuracy. Current error is, summed over all states:

Pt = total population
Rt = total (target) reps

| Pt   Ps |
| -- - -- | * Ps
| Rt   Rs |

For each state we might give a seat to, the effect that would have on total error is:

| Pt    Ps  |        | Pt   Ps |
| -- - ---- | * Ps - | -- - -- | * Ps
| Rt   Rs+1 |        | Rt   Rs | 

We would like to identify the state that minimizes this quantity. Since we are adding representatives one by one, Pt/Rt will always be greater than Ps/Rs [1] and we can remove the absolute value and distribute the Ps.

PtPs   PsPs   PtPs   PsPs
---- - ---- - ---- + ----
 Rt    Rs+1    Rt     Rs

Cancel the PtPs/Rt and we have:

PsPs   PsPs
---- - ----
 Rs    Rs+1

Combine the two fractions and cancel again:

  PsPs
---------
Rs*(Rs+1)

Since we're trying to identify the state that minimizes the quantity, we can instead identify the state that minimizes its square root:

      Ps
---------------
sqrt(Rs*(Rs+1))

Which is in the prioritization of Huntington-Hill.

I initially tried to derive this from squared error, which did not work and ended up with an enormous amount of scribbles on paper.

[1] This is not quite true, as we get to assigning the very last representatives, but I think it still works?

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