The collection

Every essay — page 21

Page 21 of 21, continuing through the fields in the same order.

Geometry Analysis Algebra Discrete Topology Probability Number Dynamics Logic Computation Applied What's new Series Concepts Search

Applied

A rule for choosing, stated exactly, and what it forces on whoever adopts it.

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.

5 figures
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.

5 figures
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.

5 figures
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.

5 figures
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.

5 figures
Five rules, and the one condition each of them gives up. A table with one row per aggregation rule and one column per condition, marking which conditions each rule satisfies when run over every profile of the agenda.

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.

6 figures
A signal both can see, and neither wants to disobey. A two-by-two game with a distribution over its four cells, drawn as the weight on each. Obeying the recommendation is a best reply for both choosers, and the pair collects 21/2 between them.

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.

6 figures
A landscape nobody is looking at, and every move goes downhill on it. The 8 states of a congestion game with 3 participants and two resources, ordered by Rosenthal's potential, with every improving unilateral move drawn as an arrow. Every arrow points downward.

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.

7 figures
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.

8 figures
A population that settles at 2/3. A contest over a prize worth 4 that costs 6 to fight for. Left: the growth rate of the share playing Hawk against that share, which is nought at 0, at 2/3 and at 1. Right: the share over time from 5 starting points, all converging on 2/3.

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.

5 figures
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.

6 figures
Biproportional seats against their fair shares. A table of votes for 3 districts and 4 parties beside the seats the biproportional method gives, each with the fair share from the continuous fit, and the cell whose seats fall outside its quota marked.

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.

5 figures
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.

5 figures
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.

7 figures · new
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.

7 figures · new