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We construct sequencings for many groups that are a semi-direct product of an odd-order abelian group and a cyclic group of odd prime order. It follows from these constructions that there is a group-based complete Latin square of order $n$…

Combinatorics · Mathematics 2018-12-14 M. A. Ollis , Christopher R. Tripp

We consider two-dimensional lattice equations defined on an elementary square of the Cartesian lattice and depending on the variables at the corners of the quadrilateral. For such equations the property often associated with integrability…

Exactly Solvable and Integrable Systems · Physics 2019-03-12 Jarmo Hietarinta

Let $C(L)$ be the right-angled Coxeter group defined by an abstract triangulation $L$ of $\mathbb{S}^2$. We show that $C(L)$ is isomorphic to a hyperbolic right-angled reflection group if and only if $L$ can be realized as an acute…

Geometric Topology · Mathematics 2019-02-07 Sang-hyun Kim , Genevieve S. Walsh

A Latin square is reduced if its first row and column are in natural order. For Latin squares of a particular order $n$ there are four possible different parities. We confirm a conjecture of Stones and Wanless by showing asymptotic equality…

Combinatorics · Mathematics 2016-10-21 Nicholas J. Cavenagh , Ian M. Wanless

The parity type of a Latin square is defined in terms of the numbers of even and odd rows and columns. It is related to an Alon-Tarsi-like conjecture that applies to Latin squares of odd order. Parity types are used to derive upper bounds…

Combinatorics · Mathematics 2013-04-17 Daniel Kotlar

The mean king's problem with maximal mutually unbiased bases (MUB's) in general dimension d is investigated. It is shown that a solution of the problem exists if and only if the maximal number (d+1) of orthogonal Latin squares exists. This…

Quantum Physics · Physics 2009-11-11 A. Hayashi , M. Horibe , T. Hashimoto

A group-theoretical approach to the construction of quasiperiodic tilings of a Euclidean plane, possessing five-fold symmetry, is applied. Of the infinitely many of variants of quasiperiodic partitions of the plane, possessing the dihedral…

General Mathematics · Mathematics 2019-08-08 Alexander S. Prokhoda

Until now the problem counting Latin rectangles m x n has been solved with an explicit formula for m = 2, 3 and 4 only. In the present paper an explicit formula is provided for the calculation of the number of Latin rectangles for any order…

Combinatorics · Mathematics 2007-11-06 Aurelio de Gennaro

A Latin tableau of shape $\lambda$ and type $\mu$ is a Young diagram of shape $\lambda$ in which each box contains a single positive integer, with no repeated integers in any row or column, and the $i$th most common integer appearing…

Combinatorics · Mathematics 2024-08-09 Timothy Y. Chow , Mark G. Tiefenbruck

A latin bitrade $(T^{\diamond}, T^{\otimes})$ is a pair of partial latin squares which defines the difference between two arbitrary latin squares $L^{\diamond} \supseteq T^{\diamond}$ and $L^{\diamond} \supseteq T^{\otimes}$ of the same…

Combinatorics · Mathematics 2008-03-08 Carlo Hamalainen

There exists a bijection between the set of Latin squares of order $n$ and the set of feasible solutions of the 3-dimensional planar assignment problem ($3PAP_n$). In this paper, we prove that, given a Latin square isotopism $\Theta$, we…

Combinatorics · Mathematics 2011-05-06 R. M. Falcón , J. Martín-Morales

We derive necessary and sufficient conditions for there to exist a latin square of order $n$ containing two subsquares of order $a$ and $b$ that intersect in a subsquare of order $c$. We also solve the case of two disjoint subsquares. We…

Combinatorics · Mathematics 2015-09-21 Joshua M. Browning , Petr Vojtěchovský , Ian M. Wanless

We use a greedy probabilistic method to prove that for every $\epsilon > 0$, every $m\times n$ Latin rectangle on $n$ symbols has an orthogonal mate, where $m=(1-\epsilon)n$. That is, we show the existence of a second Latin rectangle such…

Combinatorics · Mathematics 2007-05-23 Roland Häggkvist , Anders Johansson

There are three generalizations of the Platonic solids that exist in all dimensions, namely the hypertetrahedron, the hypercube, and the hyperoctahedron, with the latter two being dual. Conformal field theories with the associated symmetry…

High Energy Physics - Theory · Physics 2018-06-13 Andreas Stergiou

We describe the Terwilliger algebras of the four-class Latin-square association schemes arising from Cayley tables of Bol loops. We give some necessary conditions involving Terwilliger algebras for a quasigroup to be a Bol loop.

Rings and Algebras · Mathematics 2024-05-21 Brian Curtin

In an earlier paper by three of the present authors and Csaba Schneider, it was shown that, for $m\ge2$, a set of $m+1$ partitions of a set $\Omega$, any $m$ of which are the minimal non-trivial elements of a Cartesian lattice, either form…

Combinatorics · Mathematics 2022-10-14 R. A. Bailey , Peter J. Cameron , Michael Kinyon , Cheryl E. Praeger

We introduce a condition on arrays in some way maximally distinct from Latin square condition, as well as some other conditions on algebras, graphs and $0,1$-matrices. We show that these are essentially the same structures, generalising a…

Combinatorics · Mathematics 2011-08-09 Tim Boykett

We present structures comprised of identical convex polyhedra which are interlocked geometrically. These sets cannot be disassembled by removing individual polyhedra by translations and/or rotations. The shapes that permit interlocking…

Metric Geometry · Mathematics 2017-12-05 A. J. Kanel-Belov , A. V. Dyskin , Y. Estrin , E. Pasternak , I. A. Ivanov-Pogodaev

To get another from a given latin square, we have to change at least 4 entries. We show how to find these entries and how to change them.

Combinatorics · Mathematics 2019-02-18 I. I. Deriyenko

A function $f:\{0,...,q-1\}^n\to\{0,...,q-1\}$ invertible in each argument is called a latin hypercube. A collection $(\pi_0,\pi_1,...,\pi_n)$ of permutations of $\{0,...,q-1\}$ is called an autotopism of a latin hypercube $f$ if…

Information Theory · Computer Science 2019-08-29 Denis Krotov , Vladimir Potapov