SEKE 2012 Proceedings - Knowledge Systems Institute

satisfies R( D) R( C) S 1. Compared to C, D is obtained
1
by deleting one column vector from C. Assume the deleted
column vector is the j th column of C and denote it by C .
Next we will construct a non-trivial integer solution of
A Y 0 with D, C and j defined above.
S1 S1
j
Lemma 4 For the equation A Y 0
S1 S1
, denote the k
th
element
of
YS 1
by
y
S 1 : k
. Let
y
S1 :
j
det( D)
. And for i
1,
S i j
1 ,
let y det( D ( C )) , where
S1 : i i j
det( D ) is the determinant of
D, D ( C ) denote the resulted matrix after replace the i th
i
j
column of D with vector C j . Then Y y , y ,..., y
is a non-trivial integer solution of A Y 0
S1 S1
.
C
k
S S1:1 S1:2 S1:
S
1 1
Proof : Denote the k th column vector of the matrix C by
. Denote th e resulted vector after deleting the j th element
from YS 1
by Y . According to th e relationship between D
S 1
/ j
and C, CY 0
S1
has the following equivalent transformations:
y
S1:1
S1
y
S1:2
CY 0 C , C ,..., C y C
0
S1
1 2 S
S 1
1:
i i
...
i1
y
S1:
S1
y C y C 0 DY y C
0
1,
i j j S
1,..., 1,...| | 1
DY y C
/ :
S j S j j
1
1
: i : / :
S i S j j S j S j j
1 1 1
1
Let y det( D)
, then the above equation becomes
S1 : j
DY det( D)*
C . Because D is a full-rank matrix, the
S1 / j j
solution of DY det( D)*
C is unique. According to the
S1 / j j
Cramer’s rule [10] , the i th element in the solution is as follows:
det( ( det( )* ))
D D C
1 : i
det( D )
i
j
y det( D ( C ))
i j
S
i 1, 2,..., j 1, j 1,... S .
where
1
Thus, y det( D ( C ))
S1 : i i j i 1, 2,..., j 1, j 1,... S1
constitutes a solution of DY y C
S1 / j S1:
j j
when
y det( D)
. Because DY y C is equivalently
S1 : j
S1 / j S1:
j j
transformed from CY 0 , so y det( D ( C ))
S1
S1 : i i j
i 1, 2,..., j 1, j 1,... S1
and y det( D)
constitute one
S1 : j
solution of CY 0 (i.e., Y
S1
S
is a solu tion of CY 0
1
S1
). In
addition, because A Y 0 and CY 0 are equivalent, so
S1 S1
S1
Y
S 1
is also a solution of A Y 0
S1 S1
.
Furthermore, according to th e construction method of D
and C , their elements come from . According to
j
A
S 1
j
T
Definition 1, the elements of A
S 1
are all in tegers, so D and
D ( C ) are integer matrixes. Hence y det( D)
and
i j
S1 : j
y det( D ( C )) are all in tegers. Finally, y det( D)
is
S1 : i i j
S1 : j
nonzero because D is full-ranked. Therefore, Y is a nontrivial
integer solution of A Y 0
S1
S1 S1
. ■
According to T heorem 4, the solu tion Y constructed in
S1
Lemma 4 must be positive or negative integer vector. If Y
is
S1
positive let Y Y
, else let Y 1
Y
. Then Y must be a
S1 S1
S 1 S 1
S
1
positive integer solution of A Y 0
S1 S1
. Each element of Y
S1
corresponds to a place in S 1
. For each place in S S , we add
1
a zero elem ent to Y and denote th e resulted vector by Y .
S1
Then it’s obviously that Y is an S-invariants supported by S .
1
As a result, we can get the following algorithm.
Algorithm 2 Computation of a minimal-supported S-invariant
INPUT: a minimal support of S-invariants S and its generated sub-matrix
1
OUTPUT: one integer S-invariant supported by S
1
PROCEDURES:
Step1: Get the great linearly independent group of
A
S 1
A's
S 1
row spaces with
elementary column transformation method [10] . Denote the matrix composed of
these row vectors by C;
Step2: Get the great linearly independent group of C’s column spaces
with elementary row transformation method [10] . Denote the matrix composed
of these column vectors by D;
Step3: In fact, D is obtained by deleting one column vector from C.
Assume the deleted column vector is the j th column of C and denote it by C ;
j
Step4: For equation A Y
S1
0 , denote the k
th
element of YS 1
by y . Let
S 1 : S1
y det( D)
. And for i 1,
S i j
S1 : j
1 , let y det( D ( C ))
S1 : i i j
.
T
Construct the vector Y y , y ,..., y ;
S1
S1:1 S1:2 S1:
S1
Step5: If Y
S
is positive, let Y
S
Y
S
,else let Y S
1Y
S
;
1
1 1
1 1
Step6: Denote the element of Y S 1
corresponding to place s S
i 1
by
y ( s ) .Construct an |S|-dimensional column vector Y as follows:
i S1
y
( s ) if s S
S1 j j
y
( s )
j
0 if s S
j 1
1
where y ( s j
) is the element of Y which corresponds to s S ;
j
Step7: Output that Y is an S-invariants supported by S .
1
Considering the Petri net and S s , s , s , s , s
2 3 4 5 6 7
in Fig.2.
In Step1, the great linearly independent group of A
S 2
's row
spaces constitutes C as follows.
1 1 0 0 0 1 1 0 0 0
0 1 1 1 0 0 1 1 1 0
C D C
5
0 0 1 1 0 0 0 1 1 0
0 0 0 1 1 0 0 0 1 1
In Step2, the great linearly independent group of C’s
column spaces constitute D shown above. In Step3, it’s easy to
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