Re-instate Example 5.6

[MHenderson]

\begin{example}
Continuing from before,

\begin{equation}
A = \left[\begin{array}{c|*{13}c}
 & 0_1 & 1_1 & 2_1 & 3_1 & 4_1 & 5_1 & 6_1 & 7_1 & 8_1 & 9_1 & 10_1 & 11_1 & 12_1\\ \hline
0_1 &  & 2_2 12_2 & 10_1 5_1 & 6_2 10_2 & 9_1 8_1 & 12_1 6_1 & 4_1 2_1 & 11_2 9_2 & 7_2 1_2 & 5_2 4_2 & 3_1 7_1 & 8_2 3_2 & 1_1 11_1\\
1_1 & 2_1 12_1 & & 3_2 0_2 & 11_1 6_1 & 7_2 11_2 & 10_1 9_1 & 0_1 7_1 & 5_1 3_1 & 12_2 10_2 & 8_2 2_2 & 6_2 5_2 & 4_1 8_1 & 9_2 4_2\\
2_1 & 10_2 5_2 & 3_1 0_1 & & 4_2 1_2 & 12_1 7_1 & 8_2 12_2 & 11_1 10_1 & 1_1 8_1 & 6_1 4_1 & 0_2 11_2 & 9_2 3_2 & 7_2 6_2 & 5_1 9_1\\
3_1 & 6_1 10_1 & 11_2 6_2 & 4_1 1_1 & & 5_2 2_2 & 0_1 8_1 & 9_2 0_2 & 12_1 11_1 & 2_1 9_1 & 7_1 5_1 & 1_1 12_2 & 10_2 4_2 & 8_2 7_2\\
4_1 & 9_2 8_2 & 7_1 11_1 & 12_2 7_2 & 5_1 2_1 & & 6_2 3_2 & 1_1 9_1 & 10_2 1_2 & 0_1 12_1 & 3_1 10_1 & 8_1 6_1 & 2_2 0_2 & 11_2 5_2\\
5_1 & 12_2 6_2 & 10_2 9_2 & 8_1 12_1 & 0_2 8_2 & 6_1 3_1 & & 7_1 4_1 & 2_1 10_1 & 11_2 2_2 & 1_1 0_1 & 4_1 11_1 & 9_1 7_1 & 3_2 1_2\\
6_1 & 4_2 2_2 & 0_2 7_2 & 11_2 10_2 & 9_1 0_1 & 1_2 9_2 & 7_1 4_1 & & 8_2 5_2 & 3_1 11_1 & 12_2 3_2 & 2_1 1_1 & 5_1 12_1 & 10_1 8_1\\
7_1 & 11_1 9_1 & 5_2 3_2 & 1_2 8_2 & 12_2 11_2 & 10_1 1_1 & 2_2 10_2 & 8_1 5_1 & & 9_2 6_2 & 4_1 12_1 & 0_2 4_2 & 3_1 2_1 & 6_1 0_1\\
8_1 & 7_1 1_1 & 12_1 10_1 & 6_2 4_2 & 2_2 9_2 & 0_1 12_1 & 11_2 2_2 & 3_1 11_1 & 9_1 6_1 & & 10_2 7_2 & 5_1 0_1 & 1_2 5_2 & 4_1 3_1\\
9_1 & 5_1 4_1 & 8_1 2_1 & 0_1 11_1 & 7_2 5_2 & 3_2 10_2 & 1_2 0_2 & 12_1 3_1 & 4_2 12_2 & 10_1 7_1 & & 11_2 8_2 & 6_1 1_1 & 2_2 6_2\\
10_1 & 3_2 7_2 & 6_1 5_1 & 9_1 3_1 & 1_1 12_1 & 8_2 6_2 & 4_2 11_2 & 2_2 1_2 & 0_1 4_1 & 5_2 0_2 & 11_1 8_1 & & 12_2 9_2 & 7_1 2_1\\
11_1 & 8_1 3_1 & 4_2 8_2 & 7_1 6_1 & 10_1 4_1 & 2_1 0_1 & 9_2 7_2 & 5_2 12_2 & 3_2 2_2 & 1_1 5_1 & 6_2 1_2  & 12_1 9_1 & & 0_2 10_2\\
12_1 & 1_2 11_2 & 9_1 4_1 & 5_2 9_2 & 8_1 7_1 & 11_1 5_1 & 3_1 1_1 & 10_2 8_2 & 6_2 0_2 & 4_2 3_2 & 2_1 6_1 & 7_2 2_2 & 0_1 10_1 & \\
\end{array}\right]
\end{equation}

\begin{equation}
B = \left[\begin{array}{c|*{13}c}
 & 0_2 & 1_2 & 2_2 & 3_2 & 4_2 & 5_2 & 6_2 & 7_2 & 8_2 & 9_2 & 10_2& 11_ 2& 12_2\\\hline
0_2 &  & 12_2 2_1 & 10_1 5_2 & 10_2 6_1 & 9_2 8_1 & 12_1 6_2 & 4_1 2_2 & 9_1 11_2  & 1_1 7_2 & 4_2 5_1  & 3_2 7_1 & 3_1 8_2  & 1_2 11_1\\
1_2 & 2_2 12_1 & & 0_2 3_1  & 11_1 6_2 &  11_2 7_1 & 10_2 9_1 & 0_1 7_2 & 5_1 3_2 & 10_1 12_2  & 2_1 8_2  & 5_2 6_1  & 4_2 8_1 & 4_1 9_2 \\
2_2 & 5_1 10_2  & 3_2 0_1 & & 1_2 4_1  & 12_1 7_2 & 12_2 8_1  & 11_2 10_1 & 1_1 8_2 & 6_1 4_2 & 11_1 0_2  & 3_1 9_2  & 6_2 7_1  & 5_2 9_1\\
3_2 & 6_2 10_1 &  6_1 11_2 & 4_2 1_1 & &  2_2 5_1 & 0_1 8_2 & 0_2 9_1  & 12_2 11_1 & 2_1 9_2 & 7_1 5_2 & 12_1 1_2& 4_1 10_2  & 7_2 8_1 \\
4_2 & 8_2 9_1  & 7_2 11_1 & 7_1  12_2 & 5_2 2_1 & & 3_2 6_1  & 1_1 9_2 & 1_2 10_1 & 0_2 12_1 & 3_1 10_2 & 8_1 6_2 & 0_1 2_2  & 5_1 11_2 \\
5_2 &  6_1 12_2 & 9_2 10_1  & 8_2 12_1 & 8_1 0_2  & 6_2 3_1 & & 4_2 7_1  & 2_1 10_2 & 2_2 11_1  & 1_2 0_1 & 4_1 11_2 & 9_1 7_2 & 1_1 3_2 \\
6_2 & 2_1 4_2  & 7_1 0_2  &  10_2 11_1 & 9_2 0_1 & 9_1 1_2  & 7_2 4_1 & &  5_2 8_1 & 3_1 11_2 & 3_2 12_1  & 2_2 1_1 & 5_1 12_2 & 10_1 8_2\\
7_2 & 11_1 9_2 & 3_1 5_2  & 8_1 1_2  & 11_2 12_1  & 10_2 1_1 & 10_1 2_2  & 8_2 5_1 & &  6_2 9_1 & 4_1 12_2 &  4_2 0_1 & 3_2 2_1 & 6_1 0_1\\
8_2 & 7_1 1_2 & 12_1 10_2 & 4_1 6_2  & 9_1 2_2  &12_2 0_1  & 11_2 2_1 &  11_1 3_2 & 9_2 6_1 & &  7_2 10_1 & 5_1 0_2 &  5_2 1_1 & 4_2 3_1\\
9_2 & 5_2 4_1 & 8_1 2_2 & 0_1 11_2 & 5_1 7_2  & 10_1 3_2  &  0_2 1_1 & 12_2 3_1 & 12_1 4_2  & 10_2 7_1 & &  8_2 11_1 & 6_1 1_2 &  6_2 2_1\\
10_2 &  7_2 3_1 & 6_2 5_1 & 9_1 3_2 & 1_1 12_2 & 6_1 8_2  & 11_1 4_2  &  1_2 2_1 & 0_2 4_1 & 0_1 5_2  & 11_2 8_1 & &  9_2 12_1 & 7_1 2_2\\
11_2 & 8_1 3_2 &  8_2 4_1 & 7_2 6_1 & 10_1 4_2 & 2_1 0_2 & 7_1 9_2  & 12_1 5_2  &  2_2 3_1 & 1_2 5_1 & 1_1 6_2   & 12_2 9_1 & &  10_2 0_1\\
12_2 &  11_2 1_1 & 9_1 4_2 &  9_2 5_1 & 8_2 7_1 & 11_1 5_2 & 3_1 1_2 & 8_1 10_2  & 0_1 6_2  &  3_2 4_1 & 2_2 6_1 & 2_1 7_2  & 0_2 10_1 & \\
\end{array}\right]
\end{equation}

Now, assemble the large array and apply the construction using the following transversals,
\begin{align}
  T_1 &= \{((0 + g)_2, (5 + g)_2)\, |\, g \in Z_{13}\} \\
  T_2 &= \{((0 + g)_2, (11 + g)_2)\, |\, g \in Z_{13}\}
\end{align}

The obtained array is a $\BRS(28)$.

The finished array is an $ORS$ because it contains one ordered pair corresponding to each unordered pair from the set
\begin{equation}
\{\infty _1, \infty _2, 0_1, \ldots, p - 1_1, 0_2, \ldots, p - 1_2\}
\end{equation}
with each member of the same set occurring exactly once in each row and column.
To prove that this $ORS$ is a $\BRS$ we need to show that the block design obtained from the rows in the usual way is balanced.

Consider the blocks obtained from the first $p$ rows.
We have shown already that the first row of $A$ consisted of two blocks, left ($R_1 \cup N_2$) and right ($N_1 \cup R_2$).

The last stage of the construction put the pairs $\{\infty _2, 0_1\}$ and $\{\infty _1,0_2\}$ in row $0_1$ of the finished array.

So the blocks obtained from the first $p$ rows are,
\begin{equation}
  R_1 \cup N_2 \cup \{\infty _2, \infty _1 \}
\end{equation}
\begin{equation}
  N_1 \cup R_2 \cup \{0 _1, 0 _2 \}
\end{equation}
and their translates.

By similar reasoning the blocks obtained from the next $p$ rows are,
\begin{equation}
  R_2 \cup R_1 \cup \{0 _1, \infty _2 \}
\end{equation}
\begin{equation}
  N_1 \cup N_2 \cup \{0 _2, \infty _1 \}
\end{equation}
and their translates.

Finally the row labelled $\infty$ contributes a further two blocks to the design.
$\{0_1, \infty _1\} \cup R_1 \cup N_1$, obtained from the left hand members of pairs and $\{0_2, \infty _2\} \cup R_2 \cup N_2$, obtained from the right hand members.

Fortunately, Schellenberg has shown that these blocks do indeed form a $\BIBD$ with parameters, $(2(p + 1), p + 1, p)$.

To show this we consider another result due to Bose.
He showed that the blocks
$\{\infty, x^0, x^2, \ldots, x^{q - 3}\}\{0, x^0, x^2, \ldots, x^{q - 3}\}$
form a difference system in $GF(q)$ with
$\lambda = (q - 1)/2$.
This result implies that the blocks
$\{\infty, x^1, x^3, \ldots, x^{q - 2}\}\{0, x^1, x^3, \ldots, x^{q - 2}\}$
also form a difference system with the same concurrence number.
\end{example}