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Symmetries that Latin squares inherit from 1-factorizations. (English) Zbl 1067.05061
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.

##### MSC:
 05C70 Edge subsets with special properties (factorization, matching, partitioning, covering and packing, etc.) 05B15 Orthogonal arrays, Latin squares, Room squares 05-04 Software, source code, etc. for problems pertaining to combinatorics
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