A Latin square of order is an grid of cells, each of which contains one of distinct symbols, such that no symbol appears twice in any row or column. There is a natural correspondence between Latin squares of order and partitions of into triangles. We identify the three vertex classes of with the rows, columns and symbols of the Latin square. A triangle corresponds to the symbol appearing in the intersection of row and column of the Latin square. Since each cell contains exactly one symbol, and each symbol appears exactly once in each row and each column, the triangles corresponding to a Latin square do indeed partition .
What about partitions of into ‘s? Identify the vertex classes of with rows, columns, red symbols and blue symbols. Then a corresponds to a red symbol and a blue symbol in the intersection of row and column . If we look at just the red symbols or just the blue symbols then we see a Latin square. But we also have the extra property that each pair of red and blue symbol appears in exactly one cell of the grid. Two Latin squares with this property are called orthogonal. So pairs of orthgonal Latin squares of order correspond to decompositions of into ‘s. More generally, a sequence of mutually orthogonal Latin squares of order corresponds to a partition of the complete -partite graph on vertex classes of size into ‘s.
Suppose now that we have a partial Latin square of order , that is, a partially filled in grid of cells obeying the rules for a Latin square. Can it be completed to a Latin square? In the early 1980s several researchers proved that the answer is yes provided at most cells have been filled in total. This is best possible, as if we place ‘s and a single on the main diagonal there is no legal cell in which to place the th . The same example shows that it is not enough to ask only that each row and each column contains only a small number of non-empty cells. But what if each row, column and symbol has been used at most times? Can we then complete to a Latin square? Daykin and Häggkvist conjectured that we can, provided .
What does this mean on the graph side? Let be a subgraph of obtained by deleting a set of edge-disjoint triangles such that no vertex is in more than triangles. Then should have a triangle-decomposition if . In this paper we prove that has a triangle-decomposition provided .
In fact we prove something more general. The ‘s obtained in this way have the properties that (i) each vertex has the same number of neighbours in each other vertex class and (ii) each vertex sees at least a proportion of the vertices in each other class (a partite minimum degree condition). We prove that all such graphs have triangle-decompositions when .
The proof is based on that of the similar result for non-partite graphs in Edge decompositions of graphs with high minimum degree. However, it is not simply a translation of that proof to the partite setting. In the partite case we not only have to ensure that all of our gadgets can be embedded in partite graphs, we must also take care to ensure a stronger notion of divisibility is preserved throughout our decomposition process. This makes the proof extremely technical.
We also prove the analogous result for -decompositions of complete -partite graphs, with the less impressive bound . The connection to mutually orthogonal Latin squares is more complicated for , as the partially filled in cells only correspond to ‘s in the case where each non-empty cell contains one of each symbol, but we still show that there is some such that if each row, column and coloured symbol is used at most times then partial mutually orthogonal Latin squares can be complete.
As for the non-partite case, the bounds on are currently limited by available fractional or approximate decomposition results. Improvements to these would lead automatically to improvements of the bounds in this paper.