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