On the completion of latin rectangles to symmetric latin squares

Darryn Bryant, C. A. Rodger
2004 Journal of the Australian Mathematical Society  
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)
more » ... 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