Ben Barber, Daniela Kühn, Allan Lo, Deryk Osthus and Amelia Taylor, Journal of Combinatorial Theory, Series A, Volume 151, October 2017, Pages 146–201 PDF
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.