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We discuss the problem of existence of latin squares without a substructure consisting of six elements $(r_1,c_2,l_3)$, $(r_2,c_3,l_1)$, $(r_3,c_1,l_2)$, $(r_2,c_1,l_3)$, $(r_3,c_2,l_1)$, $(r_1,c_3,l_2)$. Equivalently, the corresponding…

Combinatorics · Mathematics 2026-01-27 Aleksandr D. Krotov , Denis S. Krotov

An arrangement of s elements in s rows and s columns, such that no element repeats more than once in each row and each column is called a Latin square of order s. If two Latin squares of the same order superimposed one on the other and in…

Discrete Mathematics · Computer Science 2011-11-09 R. N. Mohan , Moon Ho Lee , Subash Pokreal

A Latin square of order $n$ is an $n\times n$ matrix in which each row and column contains each of $n$ symbols exactly once. For $\epsilon>0$, we show that with high probability a uniformly random Latin square of order $n$ has no proper…

Combinatorics · Mathematics 2024-05-08 Michael J. Gill , Adam Mammoliti , Ian M. Wanless

This is a translation of Leonhard Euler's ``De quadratis magicis'' . It is E795 in the Enestrom index. This paper studies how to construct magic squares with certain numbers of cells, in particular 9, 16, 25 and 36. It considers some…

Combinatorics · Mathematics 2007-05-23 Leonhard Euler

It is proved that for any prescribed orientation of the triples of either a Steiner triple system or a Latin square of odd order, there exists an embedding in an orientable surface with the triples forming triangular faces and one extra…

Combinatorics · Mathematics 2019-11-19 Terry S. Griggs , Thomas A. McCourt , Jozef Siran

We introduce a notion of parity for transversals, and use it to show that in Latin squares of order $2 \bmod 4$, the number of transversals is a multiple of 4. We also demonstrate a number of relationships (mostly congruences modulo 4)…

Combinatorics · Mathematics 2020-04-30 Darcy Best , Ian M. Wanless

In this paper, we first present the relation between a transversal in a Latin square with some concepts in its Latin square graph, and give an equivalent condition for a Latin square has an orthogonal mate. The most famous open problem…

Combinatorics · Mathematics 2018-08-17 Adel P. Kazemi , Behnaz Pahlavsay

A (partial) Latin square is a table of multiplication of a (partial) quasigroup. Multiplication of a (partial) quasigroup may be considered as a set of triples. We give a necessary and sufficient condition when a set of triples is a…

Combinatorics · Mathematics 2007-05-23 L. Yu. Glebsky , C. J. Rubio

We recall the Alon-Tarsi conjecture on the number of even latin squares. We introduce a map which switches the parity of a latin square under certain requirements. An example is included.

Combinatorics · Mathematics 2025-03-05 Carolin Hannusch

In a Latin square, every row can be interpreted as a permutation, and therefore has a parity (even or odd). We prove that in a uniformly random $n\times n$ Latin square, the $n$ row parities are very well approximated by a sequence of $n$…

Probability · Mathematics 2025-09-19 Matthew Kwan , Kalina Petrova , Mehtaab Sawhney

A partial transversal $T$ of a Latin square $L$ is a set of entries of $L$ in which each row, column and symbol is represented at most once. A partial transversal is maximal if it is not contained in a larger partial transversal. Any…

Combinatorics · Mathematics 2021-03-02 Anthony B. Evans , Adam Mammoliti , Ian Wanless

Gy\'{a}rf\'{a}s and S\'{a}rk\"{o}zy conjectured that every $n\times n$ Latin square has a `cycle-free' partial transversal of size $n-2$. We confirm this conjecture in a strong sense for almost all Latin squares, by showing that as $n…

Combinatorics · Mathematics 2022-04-12 Stephen Gould , Tom Kelly

An intercalate in a Latin square is a $2\times2$ Latin subsquare. Let $N$ be the number of intercalates in a uniformly random $n\times n$ Latin square. We prove that asymptotically almost surely…

Combinatorics · Mathematics 2017-01-18 Matthew Kwan , Benny Sudakov

A paper by Cavenagh and Wanless diagnosed the possible intersection of any two transversals of the back circulant Latin square B_n, and used the result to completely determine the spectrum for 2-way k-homogeneous latin trades. We give a…

Combinatorics · Mathematics 2015-03-17 Trent Gregory Marbach

A $k$-plex in a latin square of order $n$ is a selection of $kn$ entries that includes $k$ representatives from each row and column and $k$ occurrences of each symbol. A $1$-plex is also known as a transversal. It is well known that if $n$…

Combinatorics · Mathematics 2018-01-10 Nicholas J. Cavenagh , Ian M. Wanless

A \emph{Latin square} is a matrix of symbols such that each symbol occurs exactly once in each row and column. A Latin square $L$ is \emph{row-Hamiltonian} if the permutation induced by each pair of distinct rows of $L$ is a full cycle…

Combinatorics · Mathematics 2023-12-21 Jack Allsop , Ian M. Wanless

Let $L$ be an order-$n$ Latin square. For $X, Y, Z \subseteq \{1, ... ,n\}$, let $L(X, Y. Z)$ be the number of triples $i\in X, j\in Y, k\in Z$ such that $L(i,j) = k$. We conjecture that asymptotically almost every Latin square satisfies…

Combinatorics · Mathematics 2016-07-26 Nathan Linial , Zur Luria

In 2008, Cavenagh and Dr\'{a}pal, et al, described a method of constructing Latin trades using groups. The Latin trades that arise from this construction are entry-transitive (that is, there always exists an autoparatopism of the Latin…

Combinatorics · Mathematics 2023-08-30 Nicholas Cavenagh , Raúl Falcón

Goyeneche et al recently proposed a notion of orthogonality for quantum Latin squares, and showed that orthogonal quantum Latin squares yield quantum codes. We give a simplified characterization of orthogonality for quantum Latin squares,…

Quantum Physics · Physics 2019-01-30 Benjamin Musto , Jamie Vicary

Ever since E. T. Parker constructed an orthogonal pair of $10\times10$ Latin squares in 1959, an orthogonal triple of $10\times10$ Latin squares has been one of the most sought-after combinatorial designs. Despite extensive work, the…

Combinatorics · Mathematics 2026-02-17 Curtis Bright , Amadou Keita , Brett Stevens