Information Technology Reference
In-Depth Information
2.
For
i
=1
,...,m
do
If
δ
k
+1
(
M
k, i
(
z
)
,T
)=
0
p
then
set
M
k
+1
,i
(
z
)=
M
k, i
(
z
).
else
Set
δ
i
←
δ
k
+1
(
M
k, i
(
z
)
,T
),
v
i
←−
k
−
1+
v
(
M
k, i
(
z
)).
,a
p
) such that
δ
i
=
j
=1
a
j
δ
j
.
Find an integer
h
such that
v
h
=max
Find a vector (
a
1
,
···
v
j
:1
{
≤
j
≤
p, a
j
}
=0
.
v
h
then
Set
M
k
+1
,i
(
z
)=
M
k, i
(
z
)
If
v
i
≥
−
j
=1
a
j
z
−v
j
+
v
i
c
j
(
z
)
.
else
−
j
=1
a
j
z
−v
j
+
v
h
c
j
(
z
)
Set
M
k
+1
,i
(
z
)=
z
−v
i
+
v
h
M
k, i
(
z
)
and
v
h
←
v
i
,
c
h
(
z
)
←
M
k, i
(
z
),
δ
h
←
δ
i
.
end-If
end-If
end-For
3.
If
k
+1=
N
then
set
M
N
(
z
)
(
M
N,i
)
1
≤i≤m
, output pol(
T
(
z
)
M
N
(
z
))
M
−
1
N
←
(
z
)andterminate
the algorithm.
else
set
k
←
k
+1, goto 2.
end-If
5AnExamp e
We use an example in [5]. Given a matrix sequence
T
(
z
)=
−
z
−
1
+
01
00
z
−
2
+
11
−
z
−
3
+
12
01
z
−
4
+
14
11
z
−
5
+
11
00
11
27
12
z
−
6
+
···
,
the goal is to find a 6th minimal partial realization of
T
.
We use Algorithm 4.1 to compute it.
r
= 0. The initial basis is
ω
1
←
1
,
0
,
0
,
0)
t
,
ω
2
←
1
,
0
,
0)
t
,
ω
3
←
z
−
1
+
(
−
(0
,
−
(
−
z
−
3
+
z
−
4
+
z
−
5
+2
z
−
6
,
z
−
3
+
z
−
5
+
z
−
6
,z
−
7
,
0)
t
,and
ω
4
←
(
z
−
1
+
z
−
2
+
z
−
3
+
−
2
z
−
4
+4
z
−
5
+7
z
−
6
,z
−
3
+
z
−
4
+
z
−
5
+2
z
−
6
,
0
,z
−
7
)
t
.
r
=1.
I
=
.
Since
θ
(
ω
3
)=
θ
(
ω
1
)and
v
(
ω
3
)
<v
(
ω
1
), we have
ξ
{
3
,
4
}
←
zω
3
−
ω
1
=(
z
−
2
+
z
−
3
+
z
−
4
+2
z
−
5
,
z
−
2
+
z
−
4
+
z
−
5
,z
−
6
,
0)
t
,
ω
1
←
−
ω
3
,
ω
3
←
ξ
.
ω
4
+
ω
1
=(
z
−
2
+
2
z
−
3
+3
z
−
4
+5
z
−
5
+9
z
−
6
,z
−
4
+2
z
−
5
+3
z
−
6
,z
−
7
,z
−
7
)
t
, and the others are
unchanged.
Since
θ
(
ω
4
)=
−
θ
(
ω
1
)and
v
(
ω
4
)=
v
(
ω
1
), we have
ω
4
←
r
=2.
I
=
{
4
}
.
