SEKE 2012 Proceedings - Knowledge Systems Institute

support because S S S which means that S is the union
3 1 2
3
of other supports.
III.
PROPERTIES OF MINIMAL SUPPORTS OF S-INVARIANTS
In this section, we will give a suff icient and necessary
condition for a place s ubset to be a m inimal support of S-
invariants. Before that, some properties are given below.
Lemma 1 [12] If S is an support of S-invariants, but not a
0
minimal support, then there are at least t wo minimal supports
of S-invariants S , S ,..., S
1 2 k 2
k such that S S S ... S .
0 1 2 k
Lemma 2 Let C be an m*
n ordered integer matrix and the
rank of C satisfies RC
n . Then the homogeneous
equation CX 0 has a fundamental system of solution in
which all the base solutions are integer vectors.
Proof: Because RC n and the coefficient matrix C is
an integer matrix, using Gaussian elimination method, we can
get a fu ndamental system of solu tion which is co nsisted of
rational vectors for the homogeneous equation CX 0 . We
denote the system by = , ,...,
1 2 k . Because
1
is a ration al vector, so we can suppose that
i
i k
i1 i2 in
...
i
i1 i2 in
all integers and v ,..., v are nonzero.
Let v
i
n
ij
j 1
i1
and
T
in
, where ,..., and v ,..., v are
i1
n n n
'
v v v
i i i i1 ij i2
ij in
.
ij
j1 j1 j1
j1 j2
jn
'
Then it is eas y to see th at
i
is an in teger vector. And as a
constant times of the base solution vector for the
i
'
homogeneous equation CX 0 , is surely to be a base
solution of CX 0
. As a result, ' ' , ' ,...,
'
1 2 k
i
in
constitute
.■
a fundamental system of integer solution of CX 0
Definition 1 [4] Let A be an incidence matrix of a Petri net
( ST , ; FW , , M)
A 1 i S denotes the i th
, and
0
column vector in A. For a given place s ubset
S s , s ,..., s
1 j1 j2
j k
S , A A , A ,..., A
S1 j1 j2
j k , a sub-matrix
of A, is called the generated sub-matrix corresponding to S .
1
Theorem 1 Let S
1 s , s ,..., s
j1 j j k
2
i
i1
be a place su bset,
in
T
A
S 1
be
the generated sub-matrix corresponding to S .If the rank of
1
A
S 1
satisfies R( A ) S 1
S1 1
and A Y 0 has positive
S1 S1
integer solutions, then S is a minimal support of S-invariants.
1
Proof: Since
A Y
S1 S1
has positive integer solutions, S1
0
must be a su pport of S -invariants. Assuming that S is not a
1
minimal support of S-invariants, then there are at leas t two
minimal supports of S-invariants S , S ,..., S ( k 2) such
11 12 1k
that S S S ... S according to Lemma 1. Let Y
1 11 12 1k
i
and Y (
j
1 i,
jk i j ) be the minimal S-invariants
supported by S and S respectively. It’s obviously that
1i
1 j
Y i
and Y are linearly independent and AY AY 0
j
i j .
Denote the l th element of Y by Y
i
i l . If Yi
l
's
corresponding place s does not belong to S , then Y
l
1
l
0
i
because the support of S-invariant Y , i.e. S , is a su bset of
i
1i
S . Deleting all those elements whose corresponding places
1
don’t belong to S from Y , and denoting the resulted vector
1
i
by , then A 0 holds because
i
S1
i
is constructed by
i
deleting some zero ele ments from Y and AY 0
i
i
holds.Similarly, deleting all tho se elements whose
corresponding places don’t belong to S from Y , and
1
j
denoting the resulted vector by , then the equation
j
A 0 holds.
S1
j
In addition, because Y and Y are linearly independent,
i
j
i
and are obtained by deleting the same zero elements from
j
Y and Y respectively, so
i
j
and must be linearly
i
j
independent too. F urthermore, from A 0
S1
i and A 0
S1
j
we can see and are two linear independent solutions of
i
j
A Y 0
S1 S1
. Only when R( A ) S 2
S1 1
, A Y 0
S1 S1
can have
such two independent solutions. But this is contradictory with
the assumption R( A ) S 1
S1 1
in the theorem. So S must be
1
a minimal support of S-invariants. ■
s 1
t 5 t 6
t 7
t 4
t 2
s 2
s 5
s 3
s 4
s 7 t 8
t 3
s 6
t 1
Fig 2. A Petri net example 2
Now let’s verify Theorem 1 w ith Peri n et in Fig.2. In
2
fact, has two minimal supports of S-invariants, which are
2
S s , s , s , s , s S s , s , s , s , s respectively.
and
1 1 2 5 6 7
2 3 4
The generated sub-matrix
S s , s , s , s , s is as follows.
1 1 2 5 6 7
5 6 7
A
S 1
corresponding to
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