TITLE: Does $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs?
QUESTION [5 upvotes]: Q: Does the complete bipartite graph $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs, where $I$ is a $1$-factor (i.e., a perfect matching)?

The obvious necessary conditions work:

$K_{12,12}$ has $12^2$ edges which is divisible by $12$, the number of edges in $K_{4,4}-I$.
Vertices in $K_{12,12}$ have degree $12$ which is divisible by $3$, the degree of the vertices in $K_{4,4}-I$.

I haven't put too much effort into solving this thus far.  It seems like it would require considerable effort to solve computationally, so I'm hoping there is a cleverer solution.
An alternative way of thinking about this problem:

Q: Does there exist a $12 \times 12$ matrix containing $12$ copies of each symbol in $\{1,2,\ldots,12\}$ such that each row and each column containing a copy of symbol $i$ contains exactly $3$ copies of $i$?

This question is a specific instance of another problem I'm thinking about.

REPLY [2 votes]: Actually, there seems to be an easy construction that I overlooked when writing my question:

$K_{6,6}$ decomposes into three $K_{4,4}-I$ subgraphs (illustrated below).
We can combine four of these to decompose $K_{12,12}$ into $K_{4,4}-I$ subgraphs.

The non-edges happen to be at just the right spots for this to work:

Here's the matrix version:
$$
\begin{array}{|cccccc|}
\hline
2 & 1 & 1 & 1 & 2 & 2 \\
1 & 2 & 1 & 1 & 2 & 2 \\
1 & 1 & 1 & 3 & 3 & 3 \\
1 & 1 & 3 & 1 & 3 & 3 \\
2 & 2 & 3 & 3 & 3 & 2 \\
2 & 2 & 3 & 3 & 2 & 3 \\
\hline
\end{array}
\longrightarrow
\begin{array}{|cccccc|cccccc|}
\hline
2 & 1 & 1 & 1 & 2 & 2 & 8 & 7 & 7 & 7 & 8 & 8 \\ 1 & 2 & 1 & 1 & 2 & 2 & 7 & 8 & 7 & 7 & 8 & 8 \\ 1 & 1 & 1 & 3 & 3 & 
3 & 7 & 7 & 7 & 9 & 9 & 9 \\ 1 & 1 & 3 & 1 & 3 & 3 & 7 & 7 & 9 & 7 & 9 & 9 \\ 2 & 2 & 3 & 3 & 3 & 2 & 8 & 8 & 9 & 9 & 9 & 8 \\ 2 & 
2 & 3 & 3 & 2 & 3 & 8 & 8 & 9 & 9 & 8 & 9 \\ \hline 5 & 4 & 4 & 4 & 5 & 5 & 11 & 10 & 10 & 10 & 11 & 11 \\ 4 & 5 & 4 & 4 & 5 & 5 & 10 & 11 & 
10 & 10 & 11 & 11 \\ 4 & 4 & 4 & 6 & 6 & 6 & 10 & 10 & 10 & 12 & 12 & 12 \\ 4 & 4 & 6 & 4 & 6 & 6 & 10 & 10 & 12 & 10 & 12 & 12 \\ 
5 & 5 & 6 & 6 & 6 & 5 & 11 & 11 & 12 & 12 & 12 & 11 \\ 5 & 5 & 6 & 6 & 5 & 6 & 11 & 11 & 12 & 12 & 11 & 12 \\
\hline
\end{array}
$$