The collection

Every essay — page 29

Page 29 of 29, 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.

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.

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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.

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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.

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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.

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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.

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Even and Paz's halving, for 4 people. The recursive halving procedure run on a cake valued differently by 4 people: each round's marks and cuts, and the final pieces, each worth at least a 1/4 share to its owner. 8 marks are made.

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.

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Maximising the product of two people's values. The frontier of value pairs from dividing 4 goods between two people, with the points maximising the product, the sum and the smaller value. The product's maximum is (65.0, 54.2) and is envy-free.

The product that makes a division fair

Divide goods to make the total happiness as large as possible and the result can be monstrously unfair; make the least happy person as happy as possible and it can waste. Multiply the people's values together and maximise the product instead, and something unexpected happens — nobody envies anybody when goods can be split, and nobody envies by more than one item when they cannot.

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Every stable matching leaves out the same people. 4 stable matchings of a market with short lists, drawn as two columns joined by edges; every one leaves the same letter and number unmatched.

The people every stable answer leaves out

Let the lists be short and let one side take several partners. Stable matchings still exist and there can be many of them — but every one leaves out exactly the same people, and a member who is left with an empty place holds exactly the same partners in every one. A three-line count proves it.

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The odd ring that forbids a stable pairing. The ranked lists of 6 people and a stable partition of them drawn on a circle: pairs as plain chords, a ring of three or more as arrows from each person to the one they hold. An odd ring is present, as it is in every stable partition of this instance.

A ring that no pairing can break

Put everybody in one pool and a stable pairing may not exist. Allow rings as well as pairs and something stable always exists — and the pairs-only answer fails exactly when that stable arrangement contains a ring of odd length. Two sides make every ring even, which is the whole reason the two-sided theorem holds.

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The duality theorem's four cases, counted over 6,561 small programs. A three-by-three table crossing the status of a linear program — optimal, unbounded or infeasible — with the status of its dual, counting every small program with coefficients from minus one to one. Five of the nine cells are empty.

When one of the two numbers is missing

The duality theorem is usually quoted as an equality: a linear program and its dual reach the same number. That is one of four cases. A program can run away to infinity, or have no feasible point at all, and then its dual is forced into a matching failure. Every small program with coefficients from minus one to one has been classified, and the table has exactly four occupied cells out of nine.

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The optimum as the lowest of 3 lines, one per dual vertex. The optimal value of a linear program plotted against one right-hand side, drawn over a family of straight lines, one for each vertex of the dual feasible region. The optimum follows the lowest line throughout.

The lines the optimum lies under

Change the resources a linear program is given and its best value changes too, tracing a graph. Every solution of the dual is a straight line lying above that graph, and the graph is exactly the lowest of those lines — a bent roof of finitely many planks. Require the answer to be in whole numbers and the roof stays where it was while the graph falls away beneath it in steps, and the space between is the part of the problem no price can see.

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The nearest consistent verdict: a 3-way tie at distance 4. A table of the 4 consistent judgement sets on the agenda p, q, and p and q, each with its number of disagreements with each of 3 judges and the total; the smallest total is marked.

The nearest consistent verdict

When a court's majorities contradict each other, one repair is to announce the consistent verdict that disagrees with the judges least. It treats the premises and the conclusion alike, which neither of the two standard procedures does. On the classic case it returns a three-way tie; on five judges, with every question weighted equally, it never returns a single answer on a troubled profile at all — and what breaks the tie is a decision about which question matters more.

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Which agendas majority can vote on safely. A table of 7 agendas with the size of their largest minimally inconsistent set and the count of inconsistent majority outcomes over all profiles of three and five judges.

Agendas that cannot contradict themselves

A court voting on two unconnected questions never contradicts itself, and neither does one voting on a chain of thresholds. A court voting on two premises and their conjunction sometimes does. What separates them is the size of the smallest sets of judgements that cannot all be true: pairs are harmless, because two majorities always share a judge, and triples are not. The same count says exactly how large a supermajority has to be to stay consistent on any agenda.

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Credit for one prediction, three ways of leaving an input out. Groups of bars, one group for each way of filling in the inputs that are not known, each bar one input's share of the difference between the prediction and the starting value.

What a missing input is worth

A model prices a house at 180 from its size, its garden and its bedrooms, and the question is how much of the price each input is responsible for. Make the inputs the players and the average over orders answers it — once somebody decides what the model says when an input is not known. Three reasonable decisions give bedrooms nothing, nothing, and sixteen, for a model that never reads them.

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Shares on two triangles and a go-between. A network of players with each node labelled by its share of what the whole network earns, and its number of links beneath it.

Cutting a link costs both of its ends the same

Three players, any two of whom can earn 1 together — but only if they are linked. Link all three and each is due a third. Remove one link and the player holding both of the others is due two thirds. Averaging over orders on the game the network allows is the one rule under which breaking any link costs the two players it joined exactly the same, and it pays go-betweens more than their links.

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How often three candidates' majorities go in a circle. The exact probability that pairwise majority among three candidates is cyclic, for every odd number of voters from 1 to 41, under two models of how ballots are drawn, with their limits of about 8.77% and 6.25%.

How often the majority goes in a circle

Three voters and three candidates give 216 profiles, and 12 of them are cycles. Count every electorate up to 41 voters exactly and the share climbs towards 8.77%, a number Guilbaud found in 1952 as the solid angle where three half-spaces at the tetrahedral angle overlap. Add candidates and a winner goes missing half the time; let voters share one axis and cycles vanish. The number is always a property of the model of how ballots are drawn.

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A majority of independent voters, more often right than any of them. The probability that a simple majority of n independent voters is right, for n from 1 to 201, when each voter is right with probability 0.45, 0.51, 0.55, 0.6, 0.7.

A majority wiser than its members

Condorcet's other theorem turns voting round: the voters no longer have preferences but judgements about a single fact, each a little more likely right than wrong. Then a simple majority of many of them is almost certainly right — 6,763 voters who are each right 51% of the time make a majority right 95% of the time. The theorem survives voters worse than a coin, if the average is better. It does not survive voters who share their mistakes, and when their skills differ the right rule weighs votes rather than counting them.

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Who does well in a random market, as it grows. A log-log plot of the average rank of partner for the proposing side and the receiving side of random balanced markets against the market's size, with dashed curves for ln n and n over ln n.

One extra person on one side

In a random market of a thousand a side, whoever proposes gets about their seventh choice and whoever receives gets about their hundred-and-fortieth. Add one person to one side and the advantage of proposing all but disappears: the shorter side does well and the longer side badly, whichever side proposes, and most people are left with exactly one stable partner.

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15 stable matchings, by what each side pays. A scatter plot of every stable matching of one instance by the total rank each side receives, running from side one's best matching to side two's, with the median, the least-total and the most even matchings marked.

The matching in the middle

List every stable matching of a market, give each member their stable partners sorted from best to worst, and hand each the one in the middle. Nothing says the result should even be a matching — two people might pick the same partner — and yet it always is one, it is always stable, and the other side gets its median partners too.

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Envy-free up to one item, and up to any: three people, five items. A value matrix for three people, five items with two allocations beneath it: one envy-free up to one item but not up to any, and one envy-free; the counts over all 243 allocations beside it.

Envy that any single item would cure

Envy-free up to one item lets a person's envy be excused if removing the envied bundle's best item would cure it. The stronger standard asks that removing any item would — even the one that person values least. Every allocation of three people's items can be searched, and an allocation meeting the stronger standard was there every time; for two people cut and choose finds one, for three it took until 2020 to prove, and for four nobody knows.

6 figures · new
Dividing a rent of 90 three ways, on a 9-step grid. A triangle of possible rent splits, triangulated into 81 small triangles, with each grid point coloured by the room its housemate would pick; 3 small triangles have all three rooms.

A rent nobody envies

Three housemates, three rooms that are not alike, one rent. Every way of splitting the rent is a point of a triangle; ask, at each point of a fine grid, which room one housemate would take at those prices, taking turns so that each small triangle has one corner for each of them. Sperner's lemma then promises a small triangle where all three would choose different rooms — and as the grid is refined, the envy at that triangle shrinks to nothing.

6 figures · new