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On the completion of latin rectangles to symmetric latin squares

Darryn Bryant, C. A. Rodger

2004
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Journal of the Australian Mathematical Society
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We find necessary and sufficient conditions for completing an arbitrary 2 by n latin rectangle to an n by n symmetric latin square, for completing an arbitrary 2 by n latin rectangle to an n by n unipotent symmetric latin square, and for completing an arbitrary 1 by n latin rectangle to an n by n idempotent symmetric latin square. Equivalently, we prove necessary and sufficient conditions for the existence of an (n -l)-edge colouring of K n (n even), and for an n-edge colouring of K n (n odd)
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... which the colours assigned to the edges incident with two vertices are specified in advance. 2000 Mathematics subject classification: primary 05B15. Darryn Bryant and C. A. Rodger [2] n by n latin square in which the first m rows are filled and the remaining cells are empty. An easy corollary of Hall's Theorem is that any m by n latin rectangle can be completed to an n by n latin square. A latin square is symmetric if for i ^ j , the symbol in cell (i, j) is also in cell (j, i). Consider the more difficult problem of completing latin rectangles to symmetric latin squares. If m = 1, it is trivial to complete any m by n latin rectangle to a symmetric latin square. Just take any n by n symmetric latin square and permute the symbols so that the first row is the 1 by n latin rectangle. However, the question is far from trivial for m > 1. In this paper we prove necessary and sufficient conditions under which a 2 by n latin rectangle can be completed to a symmetric latin square; see Theorem 3.3. A latin square is unipotent if every diagonal cell contains the same symbol. If there exists an n by n unipotent symmetric latin square, then n is necessarily even (since the number of occurrences of each symbol, other than the symbol which occurs on the diagonal, must be even). A latin square is idempotent if symbol i is in cell (i, i) for all i. If there exists an n by n idempotent symmetric latin square, then n is necessarily odd (since each symbol occurs once on the diagonal and an even number of times not on the diagonal). We also prove necessary and sufficient conditions under which a 2 by n latin rectangle can be completed to a unipotent symmetric latin square (see Theorem 3.1), and necessary and sufficient conditions under which a 1 by n latin rectangle can be completed to an idempotent symmetric latin square (see Theorem 3.2). These results extend a rich literature of related embedding results. In the containing n by n latin square L, when either symmetry has been required of L, or the diagonal of L has been prescribed in some way, then results in the literature have limited the filled cells to all occur in cells (i,j), where i,j < r for some r < n. For example, Cruse has settled such an embedding problem in the case where L is required to be symmetric [3] . It turns out that if n is odd then because of the limitation on where the filled cells occur, this result also solves the embedding problem in the case where L is both symmetric and idempotent. The more general result that predetermines the diagonal of L when L is symmetric was later solved by Andersen [1]. The extremely difficult related problem of specifying the diagonal of L, but not requiring L to be symmetric, has yet to be solved. However, it has been solved by Rodger in the case where n > 2r + 1 [6], and in the idempotent case where n = 2r [5] . In none of the literature to this point has any result been able to deal with the case where the non-diagonal filled cells span the rows or columns of the containing latin square L, and either L is required to be symmetric or the diagonal of L is specified in some way. Such completions are the main focus of this paper.

doi:10.1017/s1446788700008739
fatcat:znjmtjqebrbalp4zm4spxvqm54