Apportionment — the series

Hamilton's method on 27 seats and 5 regions. A worksheet of populations, exact quotas, floors, remainders and the seats Hamilton's method awards to 5 regions.

The seat that vanishes when the house grows

Twenty-seven whole seats have to be divided between five regions whose exact shares are 15.417, 7.209, 1.755, 1.431 and 1.188. Every rule for rounding those five numbers breaks something, and the instance drawn here breaks all three of the classical ways at once.

part 1 · applied

Five rules, one dial. Seats for each of 5 regions at 21 settings of the rounding threshold, with the three settings that are the named methods marked; the largest region gains and the smallest loses as the threshold rises.

Five rules and one dial

Adams, Webster and Jefferson are usually taught as three rules for rounding a share. They are one rule with a number in it, and turning that number from nought to one moves seats from the smallest region to the largest, one at a time.

part 2 · applied

Which regions each rule favours. Average seats above or below exact quota for the largest and the smallest region, under each of the five methods, over 400 generated instances.

The rule with no favourites

Over four hundred instances, Jefferson's method gives the largest region a third of a seat more than its exact share and the smallest a third of a seat less. Adams reverses both. Webster's average is a hundredth of a seat, and that is not luck.

part 3 · applied

Two out of three, and never all three

Stay inside every region's quota, never take a seat away when the house grows, never take one from a region that grew faster. Each pair is achievable. All three together are not, and the proof is that no rule anywhere manages it.

part 4 · applied

What each rule is answering. A table of five apportionments against three measures of inequality between two regions, with a tick where no transfer of a seat reduces the measure; each measure certifies exactly one of the five.

Choosing what unfair means

Ask whether moving one seat between two regions would make them more equal, and the answer depends on what "equal" is measured in. Three measures, three different answers, and each of the classical methods is the one no transfer can improve for exactly one of them.

part 5 · applied

Seats to districts and to parties at once. A 4 by 3 table of seats, with every row total and every column total prescribed. The entries come from scaling the votes by one factor per row and one per column and rounding, and all the totals come out exactly right.

Seats to parties and places at once

Seats can be given to regions in proportion to one list of populations, and no rule does it perfectly. Ask for seats to regions and to parties simultaneously and the object stops being a list — and the impossibility that closed the subject does not apply.

part 6 · applied

The table inside every quota

Give seats to districts and parties at once, and every cell of the table has a fair share it ought to round from. A table rounding every cell to its floor or its ceiling, with every total exact, always exists. The biproportional method does not always choose one: here it gives a party 2 seats where its fair share is 3.088.

part 7 · applied

Sixteen halves in a three-by-three-by-three table of seats. A three-way table of fair shares drawn as three slices, one per group, with sixteen cells holding a half and every line total, along districts, parties and groups, equal to zero or one.

Where the rounding runs out

In two dimensions a table of seats inside every fair share always exists. Add a third family of totals — every district and party split between groups — and it need not. Sixteen halves in a three-by-three-by-three table meet every total, and no whole table does it without a seat where the fair share is nothing, because the halves close a loop of seven.

part 8 · applied