The table inside every quota

Seats to parties and places at once describes the biproportional method — votes divided by one number for each district and one for each party, then rounded — and says two things about it without showing either. It says the method can put a cell outside its quota, and that examples are easy to construct. Here is one, and it was not quite easy: a search of six thousand random tables found thirty-two.

Three districts elect 4, 2 and 10 members, four parties are due 5, 6, 3 and 2 seats nationally, and the votes are these. Both lists of totals are themselves apportionments of a single list — the districts’ seats from their populations, the parties’ seats from their national votes — made by the one-dimensional rules and their one dial, so every number in the margins has already been rounded once before the table is filled in.

Party B’s cell in district 3 has a fair share a little over three seats and receives two. That is a quota broken in the plainest possible way, and it happens in a table that meets every district’s total and every party’s total exactly. The rest of this essay is about where the fair share comes from, why the method passes it by, and why it did not have to.

Where a fair share comes from #

In one dimension a region’s fair share is its population times the house size over the total population — its quota. In two dimensions there are two families of totals to answer to, and the naive share, a party’s votes as a fraction of its district’s seats, answers only to the districts: add up party B’s naive shares across the districts and the result need not be the six seats party B is due.

The share that answers to both families is found by scaling. Multiply every cell of district 1 by the number that makes the district’s row add to its 4 seats, and the same for the other districts. The party columns are now wrong, so multiply every cell of each party column by the number that fixes it. That spoils the rows slightly, so fix the rows again, and keep alternating.

The alternation converges, and quickly; Deming and Stephan introduced it in 1940 to adjust census tables to known totals, and it is still the standard way of fitting a table to its margins. What it converges to has a clean description. Every cell ends as its votes times one factor for its district and one for its party, so the ratios that compare two parties across two districts — votes for A in district 1 times votes for B in district 2, over the same with the districts swapped — are exactly what they were in the votes. The fit changes the totals and keeps every comparison of two parties in two districts. That is the sense in which it is the proportional table, and its entries are the fair shares.

They are almost never whole numbers, and a table of seats has to be. A cell’s quota is then the same as in one dimension: its fair share rounded down or rounded up, and nothing else.

The seats the method gives, and why they are right by its own lights #

The biproportional method does the scaling with rounding built in. It looks for a divisor for each district and a divisor for each party such that every cell’s votes, divided by both and rounded at the half, give a table meeting every total.

Among the 247 whole tables meeting all seven totals, the method picks one, and that choice can be checked without trusting the search that made it: divide by the published divisors and round. Balinski and Demange proved in 1989 that suitable divisors always exist for tables of this kind, and the Swiss cantons that use the method publish them beside the results for exactly this reason — a voter cannot reproduce the search, but anyone can check the arithmetic.

The divisors also say why party B loses its third seat in district 3. A divisor is shared. Party B’s divisor, 217, is the same number in every district, and it has to be large because party B is strong in districts 1 and 2 as well, where its 335 and 357 votes are already turning into two seats each against small district totals. District 3’s divisor, 1.432, is set by district 3’s ten seats and all four parties in it. Divided by both, party B’s 755 votes land at 2.430, and a cell whose quotient is 2.430 gets two seats however large its fair share is.

Where the quota went #

The fair share of party B’s cell is 3.088 and the method’s quotient for it is 2.430, and the difference is not an error in either. The two numbers answer different questions.

The fair share is computed from the votes alone, fitted to the totals as fractions. The quotient is computed after every other cell has been forced into whole seats. In district 1 party C’s 485 votes have a fair share of 1.953 and take two seats; in district 3 party D’s 377 votes have a fair share of 1.526 and take two seats; party A in district 3 takes 5 against a fair share of 4.464. Each of those rounds up, each draws on a total that party B shares, and the seat that district 3 and party B both have to give up lands in their common cell. The rounding in every other cell is paid for somewhere, and the method pays for it where the multipliers make it cheapest rather than where the fair share says it belongs.

This is the two-dimensional form of a trade the one-dimensional subject already knows. Two out of three, and never all three shows that no rule for a single list of populations both stays inside every quota and avoids the population paradox, and the divisor methods are the rules that give up quota. The biproportional method is a divisor method twice over, and it gives up quota for the same reason: what it guarantees instead are properties of the whole table — that more votes never cost a party a seat in a district, and that the seats inside any part of the table are what the method would give that part on its own.

A table inside every quota always exists #

Giving up quota is a choice, not a necessity, and in two dimensions the proof that it is a choice is short enough to draw.

Start from the fair shares, which meet every total. Any district whose row still contains a cell that is not a whole number contains at least two, because the row adds to a whole number and one fraction cannot make up the difference alone. The same is true of every party column. So a walk can leave a fractional cell along its row to a second fractional cell, turn and leave that one along its column to a third, turn again along a row, and so on — never stuck, because every row and column it enters has another fractional cell to leave by. The walk must eventually return to a row or column it has already visited, and the loop it closes alternates rows and columns, so it has an even number of cells.

Add a small amount to the first cell of the loop, subtract it from the second, add to the third, and continue round. Every row the loop passes through gains at one cell and loses the same at another, and so does every column, so every total is untouched. Choose the amount so that the first cell to reach a whole number stops exactly there: no cell passes its floor or its ceiling, and one more cell is whole. Repeat until none is left. In the drawn table six pushes do it.

The first push in the figure can be followed cell by cell. The loop runs through party A and party B in districts 1 and 2: district 1’s A share of 0.255 and district 2’s B share of 1.315 are marked for adding, and district 1’s B share of 1.598 and district 2’s A share of 0.280 for subtracting. The smallest amount that brings one of the four to a whole number is 0.280, which takes district 2’s A share to exactly 0. After the push district 1’s A share is 0.535 and its B share 1.318, district 2’s B share is 1.595 — and every row and column still adds to the same total, because each gained and lost the same 0.280.

The result keeps every cell at the floor or the ceiling of its fair share and every total exact — a table inside every quota, which always exists. It is the rounding method statisticians call controlled rounding, and underneath it is the integrality of flows: the tables with prescribed row and column totals are flows in a network of districts and parties, their corners are whole-number tables, and a fractional point inside can always be pushed to a corner without leaving the box its floors and ceilings define.

Why nothing can block the walk #

The pushing argument never fails, and it is worth seeing why from the other side, as a question about what could stop it. A rounding inside every quota decides, for each fractional cell, whether it rounds up or down, and each district and each party needs a stated number of its fractional cells rounded up — exactly the fractional parts of its row or column, which add to a whole number. So the question is whether seats can be handed out to fractional cells so that every district and every party receives its required number, each cell taking at most one.

That is a matching question between districts and parties, and Hall’s condition is what decides matching questions: an assignment fails only if some group of districts demands more than the parties touching them can supply. The fair shares themselves show that no group can: they already meet every demand with fractions, and a demand that fractions can meet leaves no bottleneck for whole numbers to hit. That is why the walk never runs out of room.

The same argument, with every total equal to one, is the reason a table of fractional shares of people and tasks is a lottery over whole assignments. Pushing along an alternating loop until a cell reaches 0 or 1 is exactly how that decomposition is carried out, one whole assignment peeled off at a time. Rounding a table of seats inside its quotas is the same move stopped as soon as every cell is whole, instead of repeated until the fractions are used up.

The margins are not the culprit #

A broken quota in a table could have been inherited from its margins, if the party totals were already out of line with the national votes. They are not, and checking it isolates the second dimension as the cause.

Party A’s votes across the three districts are $45 + 64 + 918 = 1027$, party B’s are $335 + 357 + 755 = 1447$, party C’s are $485 + 40 + 267 = 792$ and party D’s are $41 + 77 + 377 = 495$, and together they make $1027 + 1447 + 792 + 495 = 3761$. Sixteen seats shared in those proportions give quotas of 4.369, 6.156, 3.369 and 2.106. The party totals are 5, 6, 3 and 2: party A rounds up and the other three round down, and every party is inside its one-dimensional quota.

The district totals come from populations rather than votes, so there is no vote quota to check them against, but they are given and not in dispute. Everything the method did to party B in district 3 happened inside the table, after both margins were fixed and fair.

So the broken quota belongs to the table and not to either list. Webster’s rule applied to one list of numbers is the rule least biased between large and small, and it kept every party inside its share. Applied twice at once, with the divisors shared across a grid, it no longer can.

How often the two disagree #

The failure is real and it is rare. In the random family drawn it happens twice in three hundred tables, while the Webster apportionment of the party totals, one dimension down, never breaks quota at all in the same three hundred. Webster’s method is known for staying inside quota almost always for a single list; the second dimension adds the shared divisors, and it is the sharing that occasionally drags a cell outside its share.

That rate is a fact about the family chosen and nothing wider. Tables with a party very strong in one district and weak elsewhere, like party B’s pattern in the example, are more exposed; tables with parties spread evenly are less. The search that found the example, over six thousand tables, found thirty-two cases, and the example drawn is the most robust of them — its seats are furthest from any rounding tie, so no small change in the votes would undo the verdict.

What the choice comes to #

There are, in the example, twelve tables inside every quota and one biproportional table outside it. Choosing one of the twelve would honour every cell’s fair share, and would need a rule for choosing among twelve that says nothing about multipliers.

Such a rule gives up what the multipliers bought. Balinski and Demange derived the biproportional method from a short list of properties of this kind — the seats inside any part of the table are the seats that part would get on its own, and more votes never cost a seat — and showed that those properties force it. Since the method they force breaks the quota here, no rule with all of those properties can promise every quota. The choice one dimension down was between measures of unfairness; here it is between a guarantee about each cell and a guarantee about the table, and a legislature using the method has chosen the table.

That is a defensible choice and it is worth stating plainly what it costs. A party can win more votes in a district than its fair share of seats reflects, by more than a whole seat, and be told correctly that its national total required it. The seat that vanishes when the house grows is the one-dimensional version of being told that, and it was the start of the whole subject.

What the tables cannot show #

Every table here is three districts by four parties. Every whole table with the example’s totals — 247 of them — can be visited, which is what makes “the one table the method can give” a checked statement rather than a quoted one. Real tables have dozens of districts and parties and are solved by the alternating algorithm with its rounding built in, not by visiting tables.

The fair shares and the costs are floating-point numbers. The fit is run until every total is right to a billionth of a seat, and the verdict that 3.088 lies above 3 does not depend on the last digits. The choice among the 247 tables compares sums of logarithms, and the runner-up sits well behind the winner, so no rounding of those sums decides it.

And the rarity is a measurement of one family. Two in three hundred is what this seed and these ranges gave; it bounds nothing about elections, whose tables are not random.

The question it leaves: what happens with a third family of totals #

Everything here depended on the table having two families of totals. The walk through fractional cells worked because each fractional cell had exactly two directions to leave by, a row and a column, and alternating between two directions always closes an even loop.

Add a third family — seats prescribed for every district and party, and also balanced between two or three groups within every district and every party — and each cell has three directions to leave by. A loop that uses them unevenly can close after an odd number of steps, and an odd loop cannot be pushed round. Where the rounding runs out builds the smallest table where that happens, and finds that the table inside every quota, guaranteed in two dimensions, can simply fail to exist in three.

A guarantee about the table, or about each cell #

A fair share is a property of a cell and a total is a property of a row or a column, and in two dimensions both can always be honoured at once. The method in use honours the totals and a different pair of properties about the whole table, and gives up the cells’ shares to do it.

When a rounding has to satisfy several totals at once, ask first whether a rounding inside every share exists, and then whether the method chosen is obliged to find it. Here the first answer is always yes and the second is no, and the gap between them is where party B’s third seat went.