Applied — page 2

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.

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.

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.

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.

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.

The corners are whole assignments

A table of shares can be written as a lottery over whole assignments, which one worked example shows. The general statement is that the corners of the set of such tables are exactly the whole assignments, and that single fact is why the whole subject is easy.

A cheapest assignment, and the proof that it is cheapest. A 4 by 4 cost table with the cheapest assignment marked, and a row price and column price beside each. Every used cell's two prices add to its cost, and the prices total the assignment's cost.

A price for every person and task

The cheapest assignment can be found without comparing it to any other. Attach a number to each person and each task so that no pair's two numbers exceed its cost, and if the numbers add to an assignment's total, that assignment is cheapest — proved, by an argument that never mentions the alternatives.

One table of shares, two different lotteries. A doubly stochastic table decomposed into whole assignments twice, by two different orders, giving two mixtures that reconstruct the same shares.

One table, two lotteries

A table of shares says what fraction of each task each person does. It does not say how — the same table is a mixture of whole assignments in many different ways, and the differences are exactly what the people being assigned would care about.

The corner that is a half on every edge. A 3-vertex graph beside a table of the 5 corners of its matching relaxation. 4 are whole and one assigns a half to every edge.

Where the corners stop being whole

The easy theory of assignment rests on one property — the relaxation of the assignment problem has whole-numbered corners. Add a single edge that closes an odd cycle and the property fails, a corner appears with a half in every coordinate, and the problem changes character completely.

16 rules, and none that survives. A table of every systematic anonymous aggregation rule for 3 judges: one row per rule, showing the verdict it gives at each count of yes-votes, whether it decides every proposition, and whether it is consistent. No row has both.

No rule escapes the doctrinal paradox

A court whose members each hold a consistent position can reach an inconsistent verdict by majority. One such case is easy to build, which invites the hope that a better rule would avoid it — and every rule that responds to the votes at all fails somewhere.

Where the two procedures part company. A table of 3 judges' verdicts on two premises and the conclusion each is committed to, with the two majorities at the foot disagreeing about the conclusion.

Deciding the premises or the conclusion

A body that cannot be both decisive and coherent has to choose which. The two live options are to vote on the reasons and let the verdict follow, or to vote on the verdict and let the reasons look after themselves — and they reach opposite answers on exactly the profiles the impossibility identifies.

Four ways out, and what each costs

An impossibility theorem lists conditions and says no rule has them all. That leaves exactly as many escapes as there are conditions, each of them a real institution — a dictator, a two-stage procedure, a supermajority, a restricted agenda — and each escape's price can be counted rather than argued about.

A signal both can see

Two choosers who randomise privately can reach a set of outcomes that is smaller, and worse, than the set they reach when a device draws one cell and whispers each of them their half of it. Nothing is enforced and nobody is bound, and the arrangement is stable anyway.

The landscape nobody is looking at

Letting participants move one at a time to whatever is currently better can cycle forever, and on a network of congestible roads it cannot. The reason is a single number attached to each state that falls by exactly what the mover saves.

Two equilibria, and two tests that disagree. The row chooser's expected payoff from each option against the column chooser's behaviour, for a joint effort worth more than a safe one. The lines cross at 0.750, which is the mixed equilibrium and the boundary between the two basins.

Two equilibria and no way to choose

A game can have two states nobody wants to leave, one paying more than the other, and the definition of an equilibrium has nothing to say about which happens. The two standard tie-breakers disagree, and the one that wins is usually the worse.

A mixture that is a population

A mixed equilibrium between two choosers is a knife-edge nobody has a reason to stand on. Read the same mixture as a population whose shares grow with how well they do, and it becomes a point every population is carried to — or one every population circles for ever without arriving.

Announcing a mixture is worth 5/3 more than any equilibrium. The leader's payoff against the probability it announces for its first action, for a leader with a dominant action that is better off not being seen to play it, with the follower's reply switching where the follower is indifferent. The best announcement is worth 11/3; the best equilibrium of the simultaneous game is worth 2.

Worth more for being seen first

Moving first sounds like a disadvantage, since the other side gets to see the move and answer it. When the move is a mixture that is announced and believed, it is never a disadvantage, it is worth exactly nothing in a game of pure conflict, and in other games it is worth more than any equilibrium — sometimes by announcing an action that would never be played in secret.

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.

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.

Five certificates against any two of three decide. A table of every minimal balanced family on three players, what each demands of the game, and whether the grand coalition's value covers it — the complete test for whether a stable split exists.

Five weighings and the question is closed

Searching the triangle of splits can only ever fail to find a stable one, which is not the same as there being none. Weighing five families of coalitions against the whole settles the question outright — and the family that fails is the proof that nothing survives.

The splits of any two of three decide that nobody can out-argue. The triangle of all splits of a joint gain, with the splits marked at which every player's loudest complaint against every other is matched by an equally loud complaint back.

An objection one player makes to another

The core lets a coalition object to everybody at once. Narrow it to one player objecting to one other, require every such objection to be met by an equally loud one coming back, and exactly one split survives — with no dictionary order anywhere in the argument.

The payoffs repetition makes available in the prisoner's dilemma. A plot of the two choosers' average payoffs, with the stage game's four cells marked, their convex hull drawn, the two minmax values shown as lines, and the region above both shaded.

Patience instead of a contract

Commitment had to assume an announcement binds. Play the same game again tomorrow and the assumption is unnecessary — the future does the binding. What it costs is that nearly every outcome becomes an equilibrium, so a theory that could not choose between two now cannot choose between infinitely many.

The band two equilibria occupy, and the point a noisy reading leaves. A line of values of the payoff parameter with three regions marked — staying out dominant, both actions equilibria, investing dominant — and a single threshold inside the middle region.

The reading that is almost right

Every account of simultaneous choice so far has assumed the payoffs are known to both choosers and known to be known. Replace that with each chooser seeing a private reading off by a little, and a band of equilibria closes to a single point — so the assumption nobody states decides the answer.

How many cuts a fair share costs

Every person can be guaranteed a share of a cake worth at least one n-th by their own measure, and the oldest rule that does it asks about n²/2 questions. Splitting the people into halves and the cake at a median mark asks about n log n — and a theorem says nothing can ask fewer. Fairness has a price, and it can be counted.