Generator

Two orthogonal Latin squares of order 5

A generator in the computation library, called 45 times across 11 essays. Below: what it draws with nothing chosen and at each mode an essay asks for, what it checks while drawing, and everywhere it is used.

latin is one function. Everything below came out of it during this build, at parameters taken from the essays rather than invented for this page — so a figure here is the same figure a reader meets in an essay, and if the generator changes, this page changes with it.

With nothing chosen

Two orthogonal Latin squares of order 5. Two Latin squares side by side and their superposition, in which every pair of symbols appears exactly once.

The 3 mutually orthogonal squares of order 4

Why order 4 carries no more than 3 orthogonal squares

Latin squares of order 4, and the ones that are also Sudoku grids

Four clues, and the only grid they allow

The 288 smallest Sudoku grids are two grids in disguise

What it checks while it draws

Collected by running the family and recording what it asserted, not written here. The count is how many separate times the claim was put to the test while these drawings were made.

the squares for 1 and 2 produce every ordered pair once ×15
and once in each column of the square for 1 ×10
every symbol appears once in each row of the square for 1 ×10
positions 1 and 2 together name every word ×10
a field of order 4 has 3 non-zero multipliers ×8
the count at order 1 is a positive number ×8
there is a field with 4 elements only if 4 is a prime power ×8
between 1 and 3 squares ×7
order 3: the field's 2 squares give a code of length 4 with every two words differing in 3 places ×7
a symbol was found for column 1 ×6
column 1 repeats no symbol ×6
column 1 was given a symbol it could take ×6
relabelling left the pair 1, 2 orthogonal ×6
symbol 0 is missing from exactly n−k of the columns ×6
the cyclic square of order 3 has 3 transversals ×5
the relabelled square for 1 starts 0…3 along its first row ×5
each word has exactly 15 others at distance 4, as the MDS weight formula requires ×4
row 1 uses every symbol once ×4
every point lies on 4 lines, one from each class ×3
the number of rows already placed is a whole number between 1 and 4 ×3
and once in each column of the square for 2α+1 ×2
and once in each column of the square for α+1 ×2
every symbol appears once in each row of the square for 2α+1 ×2
every symbol appears once in each row of the square for α+1 ×2
the 3 available symbols are all used, so an extra square would repeat one ×2
the code has 16 words ×2
the order enumerated in full is a whole number between 1 and 4 ×2
the order of the squares is a whole number between 3 and 8 ×2
48 top halves can be finished in exactly 2 ways ×1
a square laid over itself produces only its own diagonal of pairs ×1
a top half finishes 4 ways when both boxes stack the same pairs in their columns, and 2 ways otherwise ×1
A₄: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
A₄: every row of the table is a permutation ×1
A₄: the operation is associative ×1
addition and multiplication are commutative ×1
an associative Latin square has an identity ×1
an even order from 4 to 8 ×1
an irreducible polynomial of the right degree was found ×1
an odd order from 3 to 9 ×1
and 288 of them fill every 2×2 box ×1
and 9,408 of order 6 ×1
and every grid has four or eight others that differ from it in exactly four ×1
and every table whose elements are all self-inverse has one ×1
and exactly half of the larger family does ×1
and fifty-six of order five ×1
and four have every element its own inverse ×1
and its largest partial transversal has 5 cells ×1
and nine thousand four hundred and eight of order six ×1
and no pair of points is met by two lines ×1
and none of them carries the symbol already standing in that column ×1
and once in each column of the first square ×1
and once in each column of the second square ×1
and once in each column of the square for 2α ×1
and once in each column of the square for α ×1
and once in each column of the square for α² ×1
and once in each column of the square for α²+1 ×1
and once in each column of the square for α²+α ×1
and once in each column of the square for α²+α+1 ×1
and sixteen of them associate ×1
and the other 48 in exactly 4 ×1
and three of them associate ×1
and twelve have an element of order four ×1
any two coordinates already tell every word apart ×1
any two words differ in all but at most one place — the most a code this size can manage ×1
at order 11 the formula is within five per cent ×1
D₄: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
D₄: every row of the table is a permutation ×1
D₄: the operation is associative ×1
D₅: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
D₅: every row of the table is a permutation ×1
D₅: the operation is associative ×1
D₆: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
D₆: every row of the table is a permutation ×1
D₆: the operation is associative ×1
Dic₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
Dic₃: every row of the table is a permutation ×1
Dic₃: the operation is associative ×1
each class holds n lines ×1
each column may still take n−k symbols ×1
each region has four cells ×1
each square either associates or does not ×1
every grid of the smaller family stacks matching pairs in its top boxes ×1
every line holds n of the points ×1
every non-zero element has a reciprocal ×1
every ordered pair of symbols occurs exactly once ×1
every pair of points is met by some line ×1
every set of columns can still take at least as many symbols as it has columns ×1
every square of order 5 has a transversal ×1
every square of order 6 has a partial transversal of 5 cells ×1
every symbol appears once in each row of the first square ×1
every symbol appears once in each row of the second square ×1
every symbol appears once in each row of the square for 2α ×1
every symbol appears once in each row of the square for α ×1
every symbol appears once in each row of the square for α² ×1
every symbol appears once in each row of the square for α²+1 ×1
every symbol appears once in each row of the square for α²+α ×1
every symbol appears once in each row of the square for α²+α+1 ×1
every symmetry carries a grid to a grid ×1
four of the sixteen have every element its own inverse ×1
multiplication distributes over addition ×1
multiplication is associative ×1
no table with an element of order four has a transversal ×1
no three given cells ever force a unique grid ×1
no two non-zero elements multiply to zero ×1
no two pieces share a wrapped diagonal ×1
no two standardised squares carry the same entry under the corner ×1
one shift per row already placed ×1
one to three named partitions ×1
Q₈: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
Q₈: every row of the table is a permutation ×1
Q₈: the operation is associative ×1
relabelling left the pair 1, α orthogonal ×1
relabelling left the pair 1, α+1 orthogonal ×1
relabelling left the pair α, α+1 orthogonal ×1
relabelling the symbols of a grid gives a grid ×1
S₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
S₃: every row of the table is a permutation ×1
S₃: the operation is associative ×1
sixteen tables of order four associate ×1
some four do ×1
some square of order 6 has none ×1
the 288 grids fall into exactly two classes under the symmetries ×1
the alphabet size is a whole number between 3 and 9 ×1
the box partition gives the 288 grids ×1
the broken partition admits none ×1
the cyclic square has transversals exactly at odd orders ×1
the cyclic square of even order 6 has no transversal ×1
the cyclic square of order 4 is Latin and not a grid ×1
the drawn order agrees with the table ×1
the field is closed ×1
the four self-inverse tables have the same number of transversals ×1
the grids have 96 different top halves ×1
the independence estimate lands within one per cent of the true count ×1
the largest order enumerated is a whole number between 3 and 6 ×1
the Latin square view is one of transversal, mols, bound, net, count, extend, quasi, mates, partial, ryser, growth, groups, queens ×1
the multiplier of the second square is a whole number between 1 and 7 ×1
the neighbouring orders are shown or not ×1
the new row uses every symbol once ×1
the number being factorised is a whole number between 1 and 1000000 ×1
the number of columns is a whole number between 3 and 6 ×1
the number of parallel classes drawn side by side is a whole number between 2 and 5 ×1
the number of squares drawn side by side is a whole number between 1 and 4 ×1
the order of the field the squares are built from is a whole number between 3 and 8 ×1
the order of the plane is a whole number between 2 and 4 ×1
the order of the square is a whole number between 3 and 7 ×1
the pair is built from a prime order, where the construction works ×1
the parallel classes are the rows, the columns and one per square ×1
the placements are exactly the transversals ×1
the puzzle drawn has exactly one completion among the 288 grids ×1
the quasigroup view is one of counts, tables ×1
the quoted orders are orders between 7 and 11 and larger than the ones enumerated ×1
the radius of the points is a whole number between 3 and 8 ×1
the relabelled square for α starts 0…3 along its first row ×1
the relabelled square for α+1 starts 0…3 along its first row ×1
the second square uses a multiplier that makes it a different square ×1
the size of the field is a whole number between 2 and 16 ×1
the size of the permuted set is a whole number between 1 and 8 ×1
the squares for 1 and α produce every ordered pair once ×1
the squares for 1 and α+1 produce every ordered pair once ×1
the squares for α and α+1 produce every ordered pair once ×1
the total at order four agrees with every square of order four, listed ×1
the word is a whole number between 0 and 15 ×1
there are 56 reduced Latin squares of order 5 ×1
there are 576 Latin squares of order 4 ×1
there are 576 Latin squares of order four ×1
there are four reduced squares of order four ×1
there are twelve Latin squares of order three ×1
twelve of them have an element of order four ×1
two associative squares were found to draw ×1
two different grids always differ in at least four cells ×1
two distinct positions kept ×1
with any two symbols kept, exactly one word fits ×1
with its first box fixed, the top band can be finished in 56 × 6⁶ ways ×1
with one kept, q words fit ×1
ℤ₁₀: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₁₀: every row of the table is a permutation ×1
ℤ₁₀: the operation is associative ×1
ℤ₁₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₁₂: every row of the table is a permutation ×1
ℤ₁₂: the operation is associative ×1
ℤ₂ × ℤ₂ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₂ × ℤ₂ × ℤ₂: every row of the table is a permutation ×1
ℤ₂ × ℤ₂ × ℤ₂: the operation is associative ×1
ℤ₂ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₂ × ℤ₂: every row of the table is a permutation ×1
ℤ₂ × ℤ₂: the operation is associative ×1
ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₂: every row of the table is a permutation ×1
ℤ₂: the operation is associative ×1
ℤ₃ × ℤ₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₃ × ℤ₃: every row of the table is a permutation ×1
ℤ₃ × ℤ₃: the operation is associative ×1
ℤ₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₃: every row of the table is a permutation ×1
ℤ₃: the operation is associative ×1
ℤ₄ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₄ × ℤ₂: every row of the table is a permutation ×1
ℤ₄ × ℤ₂: the operation is associative ×1
ℤ₄: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₄: every row of the table is a permutation ×1
ℤ₄: the operation is associative ×1
ℤ₅: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₅: every row of the table is a permutation ×1
ℤ₅: the operation is associative ×1
ℤ₆ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₆ × ℤ₂: every row of the table is a permutation ×1
ℤ₆ × ℤ₂: the operation is associative ×1
ℤ₆: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₆: every row of the table is a permutation ×1
ℤ₆: the operation is associative ×1
ℤ₇: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₇: every row of the table is a permutation ×1
ℤ₇: the operation is associative ×1
ℤ₈: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₈: every row of the table is a permutation ×1
ℤ₈: the operation is associative ×1
ℤ₉: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic ×1
ℤ₉: every row of the table is a permutation ×1
ℤ₉: the operation is associative ×1

Where it is called

Every figure on this list is drawn by the same rule, so a change to the rule changes all of them at once. That is why the list is published.

Computation

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.

Computation

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.

Computation

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.

Discrete

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.

Computation

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.

Computation

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.

Computation

Seven points, seven lines

A geometry with seven points, in which every two points lie on exactly one line and every two lines meet in exactly one point. There are no parallels, the whole thing is built out of the two-element field, and one of its lines has to be drawn as a circle.

Computation

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.

Computation

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.

Algebra

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.

Computation

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.

The whole library · What the figures prove