Matching
Named by 20 essays across 4 fields — each of them below, with the objects they name alongside it.
One bottleneck and nothing else
A set of jobs can be filled by distinct people unless some group of jobs has too few candidates between them — and that single obstruction is the only one there is, which is what makes the theorem worth having.
A lottery over whole assignments
A table of shares in which every person's shares add to one task and every task is exactly covered is never anything more than a mixture of whole assignments — and finding the mixture is a matter of taking one complete assignment out at a time.
The widest layer and the longest chain
Order sixteen subsets by inclusion and ask for the largest collection with no two comparable. The answer is the six subsets of size two — the widest layer — and no cleverer collection beats it. Ask instead for the fewest chains covering everything, and the answer is the same number again.
Nine thousand four hundred and eight
There are four Latin squares of order four once the first row and column are fixed, fifty-six of order five, and nine thousand four hundred and eight of order six. The exact answer is known for eleven orders and for no more — and yet a half-finished square can always be finished.
The same sum without its minus signs
Delete the signs from the determinant's sum over permutations and what is left counts things directly rather than by cancellation. It is a better count and a far worse object — because the cancellation was what made the determinant computable.
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 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.
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.
When several pairs share the roads
For one pair of places, the most that can travel between them equals the cheapest cut that separates them. Give three pairs a hub of three roads to share and the two numbers come apart: the pairs can send 3/2 between them, while separating every pair costs 2. Two pairs still meet their cut, but only by splitting units in half.
What the search has when it fails
A largest matching is easy to find and hard to certify: the claim that nothing larger exists is a claim about every arrangement not tried. The certificate turns out to be free — it is the wreckage of the search that failed.
The piece that cannot pair off
Take the sides away and the obstruction to a matching changes character completely. It is no longer a shortage of partners; it is a parity, and the quantity that measures it counts pieces of odd size rather than vertices of any size.
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.
One cell short of a transversal
A transversal of a Latin square picks one cell in every row and every column with every symbol different. The cyclic squares of even order have none, and that was settled by a parity argument centuries old. Whether every square of odd order has one is a conjecture from 1967 that nobody has proved; whether every square comes within one cell of having one was settled only in 2023, and only for squares large enough.
The streets a postman walks twice
A postman must walk every street of a district and come back. If every corner has an even number of streets, no street needs walking twice. If not, some must — and the ones repeated always join the odd corners in pairs. Pricing every way of pairing them finds the shortest round; pairing the nearest corners first does not.
Enough partners in every finite group
Hall's theorem says a matching exists unless some group has too few partners between them. On an infinite graph that is false: one vertex that can be paired with anybody, and every other with exactly one, passes the test on every finite group and has no matching at all. The failure needs a vertex with infinitely many choices. Forbid that, and the theorem comes back, by an argument about trees rather than about matchings.
Matching as they arrive
Applicants arrive one at a time, and each must be given a post or turned away on the spot. Any sensible rule matches at least half as many as the best assignment chosen with hindsight, and an adversary arranging the arrivals can hold every fixed rule to exactly half. Shuffle the posts once, at random, and give each arrival its best-ranked free post: the guarantee rises to 1 − 1/e, about 63%, and no rule of any kind can do better.
Tours within half again of the best
Nobody can find the shortest tour through many cities quickly, but a tour at most half as long again as the best can be built in a few steps: the shortest tree, a cheapest pairing of the cities where the tree branches oddly, an Euler circuit, and shortcuts. Nicos Christofides found it in 1976, and for forty-five years nobody could guarantee better. A strip of cities shows the half is really lost, and Laurence Wolsey's reading of the same argument shows it bounds the linear programme too.
Matching when the arrivals are shuffled
An adversary who chooses the order in which applicants arrive can hold any fixed matching rule to half the best. Take the order away and let it be random, and the plainest rule of all — give each arrival its first free place in a fixed list — rises to 1 − 1/e, because shuffling the arrivals turns out to be exactly the trick RANKING plays with the places, seen from the other side. Shuffle both and the guarantee rises again, to somewhere between 0.696 and 0.727, and where in that interval it lies is not known.
Signs that make a determinant count
Hall's theorem says whether a board can be covered by dominoes. How many ways it can be covered is a different question, and for a general graph a hopeless one — the count is a permanent, and nobody expects an efficient way to compute permanents. On a flat board it is a determinant. Put a minus sign on the vertical pairs in every second column and the thirty-six coverings of a four-by-four board stop cancelling and add up; the chessboard has 12,988,816. A hole in the board breaks the rule in a way a cut repairs, and on K₃,₃, which cannot be drawn flat, no choice among its 512 signings works at all.
The diamond that freezes its corners
A square board's domino tilings grow like 1.3385 choices per square. Arrange the same squares in a staircase-edged diamond and the count is exactly 2 to the n(n+1)/2 — only 1.1892 choices per square, at every size. The boundary has reached into the middle. Pick one of the diamond's tilings at random and the reason is visible: outside the circle inscribed in the diamond the dominoes are frozen into solid brickwork, and only inside it is the tiling free.
Named alongside it
The objects these essays reach for when they reach for this one.
Counting argumentAssignmentBipartite graphExhaustive searchGraphLatin squarePermanentCertificateDeficiencyDeterminantDualityExistence proof