3.1 Modulus Structure of Expanded and Compressed Sequences 75
3. Let H
c
= {(j
k
,
k
−
k
), k =1, 2, ··· ,k
c
},inwhichj
k
= i
k
+ L
k
and L
k
is defined by (3.7). Then, H
c
is an expanded mode of C = H(A, B).
With the above notations, we formulate the following theorem.
Theorem 12. If H, H
∈Hand H
= H
∨ H, then the expanded operators
induced by H, H
,H
satisfy the following relationship:
H
(A, B
)=H
c
[H(A, B),B
] . (3.17)
The proof of this theorem can be constructed using the definitions of H, H
c
, H
and notations 1–3.
Definition 15. H
c
defined in (3.17) is called the second expanded operator
of A,andj
k
= i
k
+ L
k
is called the shift of the second expanded operator.
Theorem 13. For every pair H
,H ∈H, H
= H
∨ H can be decomposed
into a second expanded operation. If both C and C
are virtual expanded se-
quences of A, then the corresponding expanded modes can be simplified so that
B
=(B, B
) is composed of virtual symbols q alone.
Example 5. Comparing the three sequences
⎧
⎪
⎨
⎪
⎩
A = (010201032213020103211022301) ,
C = (01020[210]103221302[11]010321[03]1022301) ,
C
= (01020[210]103221302[1100]010321[02231]1022301) ,
we can make the following observations:
1. Sequences C and C
are both expanded sequences of A, their lengths are
respectively: n
c
=34andn
c
= 39, and their expanded modes are stated
as follows:
H = {(5, 3), (14, 2), (20, 2)} = (0000030000000020000020000000) ,
H
= {(5, 3), (14, 4), (20, 5)} = (0000030000000040000050000000) .
2. H ≤ H
and
H
= H
− H = {(14, 2), (20, 3)} = (0000000000000020000030000000) .
C
can be seen as a second expanded partition of A as follows:
A
H
→ C
H
→ C
.
3. The shifting function of the accumulated length of (A, C)is
¯
L =(0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0) .
Therefore,
H
c
= H
− H = (000000[000]00000000[002]00000[0201]0000000)
in which all segments in the square brackets connected as 0000020201
comprise the second expanded mode of B.