Latin square
Named by 15 essays across 5 fields — each of them below, with the objects they name alongside it.
The thirty-six officers
Six regiments send six officers each, one of every rank. Arrange all thirty-six in a square so that each row and each column holds every rank once and every regiment once. Euler could not, guessed why, and was wrong about the reason.
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 field's worth of squares
Two orthogonal squares of order five are easy to stumble on. Four of them, every pair orthogonal, is not a stumble — it is one line of arithmetic over a field, and the field supplies as many as the order allows.
The plane hiding in the squares
A complete family of orthogonal squares is not a collection of squares that happen to agree nowhere. It is a geometry — a plane with n² points in which every two points lie on exactly one line — and reading it that way is how the impossible orders were found.
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.
Sixteen of five hundred and seventy-six
A Latin square is a multiplication table in which every equation has exactly one solution. Ask it to be associative as well and almost every square drops out — sixteen of the five hundred and seventy-six of order four survive, and they are the two groups.
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.
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.
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.
A round table with no couple together
Seat n couples round a table, men and women alternating, so that nobody sits beside their partner. Once the women are placed the men face a board of forbidden cells that bends round a corner — and that corner is the whole difficulty. The forbidden cells form a cycle, a count of non-adjacent points on a cycle finishes the problem, and the chance of a good seating creeps towards e^(−2) far more slowly than the hat problem reaches 1/e.
A Latin square with boxes
A finished Sudoku is a Latin square of order nine with one extra rule: each 3×3 box holds every digit once. At order four the extra rule keeps exactly half of the 576 Latin squares, the 288 survivors are two grids in disguise, and no puzzle can be pinned down by fewer than four clues. At order nine every one of those questions needed a computer, and the answers are 6.67 × 10²¹ grids, 5.47 billion essentially different ones, and seventeen clues.
Orthogonal squares are a code
Write down each cell of a set of orthogonal Latin squares as a word — its row, its column, and its entry in each square — and no two words agree in more than one place. That is not a pleasant accident of the squares. It is exactly what being Latin and being orthogonal say, it makes the list an error-correcting code as good as any code of its size can be, and the squares a field builds turn out to be a Reed–Solomon code, the one on every compact disc.
A mate is rare until order ten
Euler asked whether a Latin square can have an orthogonal mate, and for every order but two and six the answer is yes for some square. For a square chosen at random the answer is different. Exactly 6 of the 56 reduced squares of order five have a mate and none of order six does; of squares drawn at random, about one in a hundred of order seven has one, one in three hundred of order eight, one in seventy of order nine — and three in five of order ten.
Ten digits need a symmetry that does not commute
A check digit should catch one wrong digit and two neighbours swapped. Over eleven symbols a weighted sum does both; over the ten decimal digits no weighted sum can, no scheme of any shape built on adding modulo ten can, and the reason is the same as the reason Euler's thirty-six officers cannot be paraded. What works is the ten symmetries of a pentagon, which do not commute.
The best of thirty-four thousand scramblings
Verhoeff's check digit multiplies digits in the symmetry group of a pentagon after scrambling each one a different number of times, and 34,040 scramblings make it catch every swap of neighbours. Graded on the rarer errors, none of them catches everything, the best catch 86 of 90 doubled-digit slips and 848 of 900 swaps across a digit — and the scrambling Verhoeff published in 1969 is one of the forty that are best at both.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchTransversalCounting argumentOrthogonal latin squaresFinite fieldMatchingPermutationProjective planeCounterexampleExistence proofImpossibilityPermanent