SEKE 2012 Proceedings - Knowledge Systems Institute

Theorem 3 For a place subset S , it is a minimal support of S-
1
invariants if and only if R( A ) S 1 and A Y 0 has
S1 1
S1 S1
positive integer solutions.
Proof: It can be derived from Theorem 1 and Theorem 2. ■
IV. DECIDABILITY OF A MINIMAL SUPPORT OF S-INVARIANTS
For the conditions of Theorem 3, whether R( A ) S 1 is
S1 1
established or n ot can be deter mined through elementary
transformation method of matrix. But it’s difficult to decide
whether A Y 0
S1 S1
has positive integer solutions. Fortunately,
the following lemma can address the problem.
Lemma 3 Let
R AS 1
S 1
1
and be an arbitrary nontrivial
solutions of A Y 0
S1 S1
. The equation A Y 0 has
S1 S1
positive integer solutions if and only if is a positive vector
or a negative vector.
Proof: First, we prove the sufficiency. Assuming
is positive or negative, we only need to construct a positive
integer solution of A Y 0
S1 S1
.
Since R AS 1
S1
, A Y 0
S1 S1
has a fundamental system
of integer solution according to Lemma 2. Let be an integer
solution in such a system.
From
R AS 1
S 1
1
, we can know that must be
constant times of . Because is positive or negative, so is
. If is negative, 1* is a p ositive integer solution of
A Y 0
S1 S1
. If is positive, itself is a positive integer
solution of A Y 0
S1 S1
. In both cases, we construct a positi ve
integer solution of A Y 0
S1 S1
successfully.
Next, we prove the necessity. If A Y 0 has positive
S1 S1
integer solutions, let be such a positi ve integer solution.
Since
R AS 1
S 1
1
there must be a nonzero constant
such that . Because is a positive integer vector,
is positive or negative. ■
Theorem 4 Let be an a rbitrary non-trivial solution of
A Y 0
S1 S1
. S is a minimal support of S-invariants if and only
1
if R( A ) S 1 and is positive or negative.
S1 1
Proof: It can be derived from Theorem 3 and Lemma 3. ■
According to T heorem 4, we can get the following
algorithm to decide a minimal support of S-invariants.
Algorithm 1 Decidability of a minimal support of S-invariants
INPUT: a place subset S 1 and its generated sub-matrix
A
S 1
OUTPUT: S
1
is a minimal support of S-invariants or not
PROCEDURES:
Step1: Transform
A
S 1
to a ladder-type matrix and compute the rank of
1
with the elementary row transformations method [10] ;
Step2: If RA S {
S 1 1
Step3: C omputing a non-trivial solution of A Y
S
0 with Gaussian
S 1 1
elimination method [10] ;
Step4: If the solution is positive or negative, output that S is a minimal
1
support of S-invariants and exit;
}
Step5: Output S isn’t a minimal support of S-invariants.
1
We perform Algorithm 1 with S s , s , s , s , s
2 3 4 5 6 7
and its
generated sub-matrix A in Fig. 2 as input.
AS 2
S 2
First, with the elementary raw transformation method,
can be transformed to a ladder-type matrix as follows.
A S
2
0
0
1
1
0
0
0
0
0
0
1
1
1
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1
1
1
1
0 0
0
0
0
1
r 4 r 3
0
0
r 8 r 7
0 0
0 0
1
0
1
0
0
0
1
0
1
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1
1
1
0
A
S 1
0
0
0
0
0
0
1
0
Obviously, R( A ) 4 1
S 2
S holds according to the
2
adder-type matrix above. Starting from the ladder-type matrix
and using Gaussian elimination method, we can compute one
solution Y 2 2 1 1 1
T
of AY 0
S
S
2
2
. The solution is
positive, so S is a minimal support of S-invariants .
2
2
In Algorithm 1, the time complexity of Step1 is O S T
1
[10] 2
. For Step 3, it is O S 1
2
the time complexity of Algorithm 1 is 1
[10] . For Step 4, it is O S
1 . So
O S T .
V. COMPUTATION OF A MINIMAL-SUPPORTED S-INVARIANT
According to Theorem 4, if S is a minimal support of S-
1
invariants, then R AS 1
S 1
1
. In this case, the great linearly
independent group of A 's row spaces must include S 1
S1
1
row vectors [10] . Denote the matrix composed of these row
vectors by C. Then C must be a S 1
1 *
S1
ordered fullrow-rank
matrix which satisfies R( C) R( A ) S 1 . And
S1 1
the equation CY 0
S1
is equivalent to A Y 0
S1 S1
because the
relationship between C and A [10] S 1
. Since RC ( ) S 1 the
1
great linearly independent group of C 's column spaces must
include S 1 column vectors. Denote the matrix composed
1
of these column vectors by D. Then D must be an
S 1 * S 1 ordered full-rank square matrix which
1 1
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