8.3 Linearly Independent Latin Arrays 345
10, 5, 4, 3, 6, 11, 8, 1, 9, 2, 7;
10, 5, 4, 3, 6, 11, 8, 1, 9, 7, 2;
10, 5, 7, 3, 6, 11, 8, 1, 9, 2, 4;
10, 5, 7, 3, 6, 11, 8, 1, 9, 4, 2;
10, 5, 4, 3, 8, 11, 6, 1, 9
, 2, 7;
10, 5, 4, 3, 8, 11, 6, 1, 9, 7, 2;
10, 5, 7, 3, 8, 11, 6, 1, 9, 2, 4;
10, 5, 7, 3, 8, 11, 6, 1, 9, 4, 2.
For any solution π in the example, let
V
0
=
0000
,V
1
=
⎡
⎢
⎢
⎣
1000
0100
0010
1100
⎤
⎥
⎥
⎦
,
and P
R
be the 11 × 11 permutation matrix of which the element at row i
column π(i) is 1. From Theorem 8.3.5,
⎡
⎣
V
0
V
1
WV
1
⊕ P
R
(I
V
0
⊕ U
−
)
⎤
⎦
is in S(V
0
,V
1
), where WV
1
= A, U
−
= B,androwsofW
i
in W are arranged
in increasing order, that is,
W =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0011
0101
0110
1001
1010
1100
0111
1011
1101
1110
1111
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
Therefore,
P
−1
⎡
⎣
V
0
V
1
WV
1
⊕ P
R
(I
V
0
⊕ U
−
)
⎤
⎦
=
⎡
⎣
0
V
1
P
R
B
⎤
⎦
gives the last 4 columns of the truth table of a permutation ϕ on R
4
2
with
c
ϕ
> 1.