an:02173304
Zbl 1067.05061
Wanless, Ian M.; Ihrig, Edwin C.
Symmetries that Latin squares inherit from 1-factorizations
EN
J. Comb. Des. 13, No. 3, 157-172 (2005).
00115717
2005
j
05C70 05B15 05-04
perfect 1-factorization; autotopy; paratopy; main class; totally symmetric; idempotent; atomic Latin square
A rooted 1-factorization, \((F,v)\), of the complete graph \(K_{n+1}\), with \(n\) odd, consists of a 1-factorization \(F\) of \(K_{n+1}\) along with a vertex \(v\) from \(K_{n+1}\), called the root, and a total order on the vertices of \(K_{n+1}\). The authors make use of the well-known \(\mathbb K\)-construction to obtain an ordered 1-factorization of the complete bipartite graph \(K_{n,n}\) and then a Latin square of order \(n\) denoted by \({\mathcal L}(F,v)\), which is shown to be symmetric and idempotent, by not necessarily totally symmetric. Conversely, any symmetric idempotent Latin square of order \(n\) is proven to be \({\mathcal L}(F,v)\) for some \((F,v)\) of \(K_{n+1}\). They establish a number of results concerning the autotopies which \({\mathcal L}(F,v)\) can possess. In particular, the automorphism group of \({\mathcal L}(F,v)\) is isomorphic to the stabilizer of \(v\) in the automorphism group of \(F\). Their main result is that the three assertions: \({\mathcal L}(F,u)\) is paratopic to \({\mathcal L}(G,v)\), \({\mathcal L}(F,u)\) is isomorphic to \({\mathcal L}(G,v)\), and \((F,u)\) is isomorphic to \((G,v)\), are equivalent. They also provide an algorithm that can determine in \(O(n^3)\) time whether a Latin square of order \(n\) is paratopic to an \({\mathcal L}(F,v)\), or whether it is isotopic to a symmetric or totally symmetric Latin square.
Chester J. Salwach (Easton)