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Fig. 2.1
Transform an
OA
(
8
,
4
·
2
4
0 0000
0 1111
1 0011
1 1100
2 0101
2 1010
3 0110
3
1001
000000
001111
010011
011100
100101
101010
110110
11
1001
,
2
)
to an
OA
(
8
,
2
6
,
2
)
by splitting
2.1.2 Splitting
(
,(
×
)
·
s
2
···
s
k
,
)
(
,
·
·
s
2
···
s
k
,
)
Theorem 2.2
If an OA
N
p
s
t
exists, then an OA
N
p
s
t
also exists.
For each element
i
∈{
0
,
1
,...,
p
×
s
−
1
}
,let
v
=
i
mod
s
,
u
=
(
i
−
v
)/
s
. Then
split the first column of OA
into two columns by replacing
element
i
with two elements
u
and
v
. The resulting array is an OA
(
N
,(
p
×
s
)
·
s
2
···
s
k
,
t
)
(
N
,
p
·
s
·
s
2
···
s
k
,
t
)
.
2
6
2
4
For example, we can construct an OA
(
8
,
,
2
)
from an OA
(
8
,
4
·
,
2
)
by splitting
the first column, as illustrated in Fig.
2.1
.
2.1.3 Hadamard Construction
-matrix
H
n
is an Hadamard matrix if
H
n
H
n
An
n
×
n
(
−
1
,
1
)
=
nI
.
Example 2.1
The matrix
⎛
⎞
1111
11
⎝
⎠
−
1
−
1
H
4
=
1
−
11
−
1
1
−
1
−
11
is an Hadamard matrix because:
⎛
⎝
⎞
⎠
×
⎛
⎝
⎞
⎠
=
⎛
⎝
⎞
⎠
.
1111
11
1111
11
4000
0400
0040
0004
−
1
−
1
−
1
−
1
1
−
11
−
1
1
−
11
−
1
−
−
−
−
1
1
11
1
1
11
2
n
Theorem 2.3
When an Hadamard matrix of order n
≥
4
exists, an OA
(
2
n
,
,
3
)
2
2
n
−
4
4
2
and an OA
(
8
n
,
·
,
3
)
exist.
Due to space limitation, we just elaborate the first case. Suppose
H
n
is an Hadamard
matrix of order
n
(
n
H
n
−
2
n
≥
4), then
is an OA
(
2
n
,
,
3
)
. Figure
2.2
shows an
H
n
2
4
OA
constructed with the Hadamard matrix
H
4
in the above example. By
convention, we replace the element '
(
8
,
,
3
)
−
1' with '0'.
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