MHenderson
commented
6 months ago
\begin{example}
Consider the following $\BRS(8)$.
\begin{equation}
\begin{bmatrix}
\infty 0 & 26 & 45 & & 13 & & \\
& \infty 1 & 30 & 56 & & 24 & \\
& & \infty 2 & 41 & 60 & & 35 \\
46 & & & \infty 3 & 52 & 01 & \\
& 50 & & & \infty 4 & 63 & 12 \\
23 & & 61 & & & \infty 5 & 04 \\
15 & 34 & & 02 & & & \infty 6 \\
\end{bmatrix}
\end{equation}
\begin{equation}
\begin{bmatrix}
\infty 0 & & & 64 & & 32 & 51 \\
62 & \infty 1 & & & 05 & & 43 \\
54 & 03 & \infty 2 & & & 16 & \\
& 65 & 14 & \infty 3 & & & 20 \\
31 & & 06 & 25 & \infty 4 & & \\
& 42 & & 10 & 36 & \infty 5 & \\
& & 53 & & 21 & 40 & \infty 6 \\
\end{bmatrix}
\end{equation}
These BRS satisfy all three properties required by Theorem~\ref{thm:schellenberg} (we shall see why later).
Suppose we take the following two disjoint common transversals as $T_1$ (lighter grey shading) and $T_2$ (darker grey shading).
\begin{equation}
R \odot S = \begin{bmatrix}
00 & 26 & 45 & 64 & 13 & 32 & 51 \\
62 & 11 & 30 & 56 & 05 & 24 & 43 \\
54 & 03 & 22 & 41 & 60 & 16 & 35 \\
46 & 65 & 14 & 33 & 42 & 01 & 20 \\
31 & 50 & 06 & 25 & 44 & 63 & 12 \\
23 & 42 & 61 & 10 & 36 & 55 & 04 \\
15 & 34 & 53 & 02 & 21 & 40 & 66 \\
\end{bmatrix}
\end{equation}
Next we obtain $A$ from the superposition of $\hat{R}$ and
$\hat{S'}$.
\begin{equation}
A = \begin{bmatrix}
& 26 & 45 & 6'4' & 13 & 3'2' & 5'1' \\
6'2' & & 30 & 56 & 0'5' & 24 & 4'3' \\
5'4' & 0'3' & & 41 & 60 & 1'6' & 35 \\
46 & 6'5' & 1'4' & & 42 & 01 & 2'0' \\
3'1' & 50 & 0'6' & 2'5' & & 63 & 12 \\
23 & 4'2' & 61 & 1'0' & 3'6' & & 04 \\
15 & 34 & 5'3' & 02 & 2'1' & 4'0' & \\
\end{bmatrix}
\end{equation}
Then we construct $B$, and after swapping the order of certain pairs, obtain $D$.
\begin{equation}
B = \begin{bmatrix}
0'0 & 26' & 45' & 64' & 13' & 32' & 51' \\
62' & 11' & 30' & 56' & 05' & 24' & 43' \\
54' & 03' & 22' & 41' & 60' & 16' & 35' \\
46' & 65' & 14' & 33' & 52' & 01' & 20' \\
31' & 50' & 06' & 25' & 44' & 63' & 12' \\
23' & 42' & 61' & 10' & 36' & 55' & 04' \\
15' & 34' & 53' & 02' & 21' & 40' & 66' \\
\end{bmatrix}
\end{equation}
\begin{equation}
D = \begin{bmatrix}
0'0 & 6'2 & 5'4 & 64' & 3'1 & 32' & 51' \\
62' & 11' & 0'3 & 6'5 & 05' & 4'2 & 43' \\
54' & 03' & 22' & 1'4 & 0'6 & 16' & 5'3 \\
6'4 & 65' & 14' & 33' & 2'5 & 1'0 & 20' \\
31' & 0'5 & 06' & 25' & 44' & 3'6 & 2'1 \\
3'2 & 42' & 1'6 & 10' & 36' & 55' & 4'0 \\
5'1 & 4'3 & 53' & 2'0 & 21' & 40' & 66' \\
\end{bmatrix}
\end{equation}
$C$ is obtained by arranging $A$ and $D$ thus.
\begin{equation}
B = \left[\begin{array}{c|*{15}c}
0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 0' & 1' & 2' & 3' & 4' & 5' & 6' & l \\ \hline
1 & & 26 & 45 & 64' & 13 & 3'2' & 5'1' & & & & & & & & \\
2 & 6'2' & & 30 & 56' & 0'5' & 24 & 4'3' & & & & & & & & \\
3 & 5'4' & 0'3' & & 41' & 60 & 1'6' & 35 & & & & & & & & \\
4 & 46 & 6'5' & 1'4' & & 52 & 01 & 2'0' & & & & & & & & \\
5 & 3'1' & 50 & 0'6' & 25' & & 63 & 12 & & & & & & & & \\
6 & 23 & 4'2' & 61 & 10' & 3'6' & & 04 & & & & & & & & \\
0' & 15 & 34 & 5'3' & 02' & 2'1' & 4'0' & & & & & & & & & \\
1' & & & & & & & & 0'0 & 6'2 & 5'4 & 64' & 3'1 & 32' & 51' & \\
2' & & & & & & & & 62' & 11' & 0'3 & 6'5 & 05' & 4'2 & 43' & \\
3' & & & & & & & & 54' & 03' & 22' & 1'4 & 0'6 & 16' & 5'3 & \\
4' & & & & & & & & 6'4 & 65' & 14' & 33' & 2'5 & 1'0 & 20' & \\
5' & & & & & & & & 31' & 0'5 & 06' & 25' & 44' & 3'6 & 2'1 & \\
6' & & & & & & & & 3'2 & 42' & 1'6 & 10' & 36' & 55' & 4'0 & \\
7' & & & & & & & & 5'1 & 4'3 & 53' & 2'0 & 21' & 40' & 66' & \\
8' & & & & & & & & & & & & & & & \infty'\infty \\
\end{array}\right]
\end{equation}
Finally, we construct $F$ according to the two presciptions in the final part of the construction.
Which involves first replacing some of the pairs in those cells of $C$ corresponding to $D$.
Consider
\begin{equation}
T'_1 = \{(2',6'),(3',0'),(4',1'),(5',2'),(6',3'),(0',4'),(1',5')\}
\end{equation}
In $R$, the cells $(2, 6), (3, 0), (4, 1), (5, 2), (6, 3), (0, 4)$ and $(1, 5)$ were all non-empty, so $C$ contains pairs of the form $(k', n)$ in all cells $(2', 6'), (3', 0'), \ldots$.
All of these are removed and replaced in the final column of the same row.
And, for each pair $(k', n)$, corresponding pairs of the form $(\infty ', k'), (\infty,n)$ are put in cells $(k, j'_m), (n, j'_m)$, respectively.
For example, $(5',3)$ appears in position $(2', 6')$ of $C$.
So in $F$, $(5', 3)$ appears in $(2', l)$ and the additional pairs $(\infty, 3)$ and $(\infty ', 5')$ appear in column $6'$, in rows 3 and 5 respectively.
All of which makes the final columns appear like so:
\begin{equation}
\left[\begin{array}{c|*{8}c}
& 0' & 1' & 2' & 3' & 4' & 5' & 6' & l \\ \hline
0 & & \infty ' 0' & & \infty 0 & & & & \\
1 & & & \infty ' 1' & & \infty 1 & & & \\
2 & & & & \infty ' 2' & & \infty 2 & & \\
3 & & & & & \infty ' 3 ' & & \infty 3 & \\
4 & \infty 4 & & & & & \infty ' 4 ' & & \\
5 & & \infty 5 & & & & & \infty ' 5 ' & \\
6 & \infty ' 6 ' & & \infty 6 & & & & & \\
0' & 00' & 6'2 & 5'4 & 64' & & 32' & 51' & 3'1 \\
1' & 62' & 11' & 0'3 & 6'5 & 05' & & 43' & 4'2 \\
2' & 54' & 03' & 22' & 1'4 & 0'6 & 16' & & 5'3 \\
3' & & 65' & 14' & 33' & 2'5 & 1'0 & 20' & 6'4 \\
4' & 31' & & 06' & 25' & 44' & 3'6 & 2'1 & 0'5 \\
5' & 3'2 & 42' & & 10' & 36' & 55' & 4'0 & 1'6 \\
6' & 5'1 & 4'3 & 53' & & 21' & 40' & 66' & 2'0 \\
l & & & & & & & & \infty ' \infty \\
\end{array}\right]
\end{equation}
Now consider:
\begin{equation}
T'_2 = \{(4', 5'), (5', 6'), (6', 0'), (0', 1'), (1', 2'), (2', 3'), (3', 4')\}
\end{equation}
In each cell $(i'_m, j'_m)$ of $C$, where $(i'_m,j'_m) \in T'_2$, occurs a pair $(k', n)$.
We remove this and place it in cell $(l, j'_m)$, also putting pairs $(k', \infty), (\infty ', n)$ in cells $(i'_m, k), (i'_m, n)$.
For example, $(3', 6)$ appears in $(4', 5')$, so $(3', 6)$ goes in $(l, 5')$ and $(3', \infty), (\infty ', 6)$ go in $(4', 3), (4', 6)$ respectively.
\begin{equation}
\left[\begin{array}{c|*{15}c}
& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 0' & 1' & 2' & 3' & 4' & 5' & 6' & l \\ \hline
0' & & & \infty ' 2 & & & & 6' \infty & 00' & & 5'4 & 64' & & 32' & 51' & 3'1 \\
1' & 0' \infty & & & \infty ' 3 & & & & 62' & 11' & & 6'5 & 05' & & 43' & 4'2 \\
2' & & 1' \infty & & & \infty ' 4 & & & 54' & 03' & 22' & & 0'6 & 16' & & 5'3 \\
3' & & & 2' \infty & & & \infty ' 5 & & & 65' & 14' & 33' & & 1'0 & 20' & 6'4 \\
4' & & & & 3' \infty & & & \infty ' 6 & 31' & & 06' & 25' & 44' & & 2'1 & 0'5 \\
5' & \infty ' 0 & & & & 4' \infty & & & 3'2 & 42' & & 10' & 36' & 55' & & 1'6 \\
6' & & \infty ' 1 & & & & 5' \infty & & & 4'3 & 53' & & 21' & 40' & 66' & 2'0 \\
l & & & & & & & & 5'1 & 6'2 & 0'3 & 1'4 & 2'5 & 3'6 & 4'0 & \infty ' \infty \\
\end{array}\right]
\end{equation}
So the following array, the completed $F$, is a $\BRS(16)$:
\begin{equation}
\left[\begin{array}{*{16}c}
& 26 & 45 & 6'4' & 13 & 3'2' & 5'1' & & \infty ' 0' & & \infty 0 & & & & \\
6'2' & & 30 & 56 & 0'5' & 24 & 4'3' & & & \infty ' 1' & & \infty 1 & & & \\
5'4' & 0'3' & & 41 & 60 & 1'6' & 35 & & & & \infty ' 2' & & \infty 2 & & \\
46 & 6'5' & 1'4' & & 52 & 01 & 2'0' & & & & & \infty ' 3' & & \infty 3 & \\
3'1' & 50 & 0'6' & 2'5' & & 63 & 12 & \infty 4 & & & & & \infty ' 4' & & \\
23 & 4'2' & 61 & 1'0' & 3'6' & & 04 & & \infty 5 & & & & & \infty ' 5' & \\
15 & 34 & 5'3' & 02 & 2'1' & 4'0' & & \infty' 6' & & \infty 6 & & & & & \\
& & \infty ' 2 & & & & 6' \infty & 00' & & 5'4 & 64' & & 32' & 51' & 3'1 \\
0' \infty & & & \infty ' 3 & & & & 62' & 11' & & 6'5 & 05' & & 43' & 4'2 \\
& 1' \infty & & & \infty ' 4 & & & 54' & 03' & 22' & & 0'6 & 16' & & 5'3 \\
& & 2' \infty & & & \infty ' 5 & & & 65' & 14' & 33' & & 1'0 & 20' & 6'4 \\
& & & 3' \infty & & & \infty ' 6 & 31' & & 06' & 25' & 44' & & 2'1 & 0'5 \\
\infty ' 0 & & & & 4' \infty & & & 3'2 & 42' & & 10' & 36' & 55' & & 1'6 \\
& \infty ' 1 & & & & 5' \infty & & & 4'3 & 53' & & 21' & 40' & 66' & 2'0 \\
& & & & & & & 5'1 & 6'2 & 0'3 & 1'4 & 2'5 & 3'6 & 4'0 & \infty ' \infty \\
\end{array}\right]
\end{equation}
\end{example}
I don't think
smallmatrixis going to save us here.