Abstract:Let q be an odd prime power and suppose that
$a,b\in \mathbb {F}_q$
are such that
$ab$
and
$(1{-}a)(1{-}b)$
are nonzero squares. Let
$Q_{a,b} = (\mathbb {F}_q,*)$
be the quasigroup in which the operation is defined by
$u*v=u+a(v{-}u)$
if
$v-u$
is a square, and
$u*v=u+b(v{-}u)$
if
$v-u$
… Show more
“…Such squares are called quadratic Latin squares. Quadratic Latin squares have previously been used to construct perfect 1-factorisations [34,15,1], mutually orthogonal Latin squares [14,13], atomic Latin squares [34], Falconer varieties [1], and maximally non-associative quasigroups [9,10]. Quadratic Latin squares are the main focus of this paper.…”
Section: Introductionmentioning
confidence: 99%
“…10. A permutation α ∈ Γ contains a Type One c-cycle if and only if it satisfies a sequence in X c,α / ∼.…”
A Latin square of order n is an n × n matrix of n symbols, such that each symbol occurs exactly once in each row and column. For an odd prime power q let F q denote the finite field of order q. A quadratic Latin square is a Latin square L[a, b] defined by,for some {a, b} ⊆ F q such that ab and (a − 1)(b − 1) are quadratic residues in F q . Quadratic Latin squares have previously been used to construct perfect 1-factorisations, mutually orthogonal Latin squares and atomic Latin squares. We first characterise quadratic Latin squares which are devoid of 2 × 2 Latin subsquares. Let G be a graph and F a 1-factorisation of G. If the union of every pair of 1-factors in F induces a Hamiltonian cycle in G then F is called perfect, and if there is no pair of 1-factors in F which induce a Hamiltonian cycle in G then F is called anti-perfect. We use quadratic Latin squares to construct new examples of anti-perfect 1-factorisations of complete graphs and complete bipartite graphs. We also demonstrate that for each odd prime p, there are only finitely many orders q, which are powers of p, such that quadratic Latin squares of order q could be used to construct perfect 1-factorisations of complete graphs or complete bipartite graphs.
“…Such squares are called quadratic Latin squares. Quadratic Latin squares have previously been used to construct perfect 1-factorisations [34,15,1], mutually orthogonal Latin squares [14,13], atomic Latin squares [34], Falconer varieties [1], and maximally non-associative quasigroups [9,10]. Quadratic Latin squares are the main focus of this paper.…”
Section: Introductionmentioning
confidence: 99%
“…10. A permutation α ∈ Γ contains a Type One c-cycle if and only if it satisfies a sequence in X c,α / ∼.…”
A Latin square of order n is an n × n matrix of n symbols, such that each symbol occurs exactly once in each row and column. For an odd prime power q let F q denote the finite field of order q. A quadratic Latin square is a Latin square L[a, b] defined by,for some {a, b} ⊆ F q such that ab and (a − 1)(b − 1) are quadratic residues in F q . Quadratic Latin squares have previously been used to construct perfect 1-factorisations, mutually orthogonal Latin squares and atomic Latin squares. We first characterise quadratic Latin squares which are devoid of 2 × 2 Latin subsquares. Let G be a graph and F a 1-factorisation of G. If the union of every pair of 1-factors in F induces a Hamiltonian cycle in G then F is called perfect, and if there is no pair of 1-factors in F which induce a Hamiltonian cycle in G then F is called anti-perfect. We use quadratic Latin squares to construct new examples of anti-perfect 1-factorisations of complete graphs and complete bipartite graphs. We also demonstrate that for each odd prime p, there are only finitely many orders q, which are powers of p, such that quadratic Latin squares of order q could be used to construct perfect 1-factorisations of complete graphs or complete bipartite graphs.
“…The condition ensures that is a Latin square [16]. Quadratic Latin squares have previously been used to construct perfect 1‐factorisations [1, 17, 38], mutually orthogonal Latin squares [15, 16], atomic Latin squares [38], Falconer varieties [1] and maximally nonassociative quasigroups [10, 11]. Quadratic Latin squares are the main focus of this paper.…”
Section: Introductionmentioning
confidence: 99%
“…[ , ] is a Latin square [16]. Quadratic Latin squares have previously been used to construct perfect 1-factorisations [1,17,38], mutually orthogonal Latin squares [15,16], atomic Latin squares [38], Falconer varieties [1] and maximally nonassociative quasigroups [10,11]. Quadratic Latin squares are the main focus of this paper.…”
Section: Introductionmentioning
confidence: 99%
“…∕ and q a ( , ) {(13, 2), (13,9), (17,5), (17,8), (37,11), (37,27), ∉ (41, 23), (41, 26)}, (iv) b a = − and q a ( , ) {(13, 7), (13,11), (17,3), (17,11), (37,10), (37,26), (41, 12), ∉ (41, 17)}.…”
A Latin square of order is an matrix of symbols, such that each symbol occurs exactly once in each row and column. For an odd prime power let denote the finite field of order . A quadratic Latin square is a Latin square defined by
for some such that and are quadratic residues in . Quadratic Latin squares have previously been used to construct perfect 1‐factorisations, mutually orthogonal Latin squares and atomic Latin squares. We first characterise quadratic Latin squares which are devoid of Latin subsquares. Let be a graph and a 1‐factorisation of . If the union of every pair of 1‐factors in induces a Hamiltonian cycle in then is called perfect, and if there is no pair of 1‐factors in which induce a Hamiltonian cycle in then is called antiperfect. We use quadratic Latin squares to construct new examples of antiperfect 1‐factorisations of complete graphs and complete bipartite graphs. We also demonstrate that for each odd prime , there are only finitely many orders , which are powers of , such that quadratic Latin squares of order could be used to construct perfect 1‐factorisations of complete graphs or complete bipartite graphs.
Let $\mathbb F$ be a finite field of odd order and $a,b\in\mathbb F\setminus\{0,1\}$ be such that $\chi(a) = \chi(b)$ and $\chi(1-a)=\chi(1-b)$, where χ is the extended quadratic character on $\mathbb F$. Let $Q_{a,b}$ be the quasigroup over $\mathbb F$ defined by $(x,y)\mapsto x+a(y-x)$ if $\chi(y-x) \geqslant 0$, and $(x,y)
\mapsto x+b(y-x)$ if $\chi(y-x) = -1$. We show that $Q_{a,b} \cong Q_{c,d}$ if and only if $\{a,b\}
= \{\alpha(c),\alpha(d)\}$ for some $\alpha\in \operatorname{Aut}(\mathbb F)$. We also characterize $\operatorname{Aut}(Q_{a,b})$ and exhibit further properties, including establishing when $Q_{a,b}$ is a Steiner quasigroup or is commutative, entropic, left or right distributive, flexible or semisymmetric. In proving our results, we also characterize the minimal subquasigroups of $Q_{a,b}$.
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