To determine the S-invariants, you have to solve a linear equation system, i.e., transpose(M)*x=0 where M is the incidence matrix. The matrix of the mentioned exercise is
2 0 0 −1 0
-5 3 1 0 0
0 −6 0 5 4
0 0 −1 0 −2 Every solution to the linear equation system is an s-invariant. You may proceed as follows using the Gaussian algorithm:
10 0 0 −5 0
-10 6 2 0 0
0 −6 0 5 4
0 0 −1 0 −2
10 0 0 −5 0
0 6 2 −5 0
0 −6 0 5 4
0 0 −1 0 −2
10 0 0 −5 0
0 6 2 −5 0
0 0 2 0 4
0 0 −1 0 −2
10 0 0 −5 0
0 6 2 −5 0
0 0 2 0 4
0 0 −2 0 −4
10 0 0 −5 0
0 6 2 −5 0
0 0 2 0 4
0 0 0 0 0
2 0 0 −1 0
0 6 0 −5 -4
0 0 2 0 4
0 0 0 0 0
From here you can get two solution vectors:
x1 = (3,5,0,6,0) x2 = (0,2,-6,0,3)
Now, any linear combination of these two vectors is an S-invariant. Hence, there are infinitely many s-invariants, and all of them can be obtained, e.g., as a linear combination of the above two ones (or the ones mentioned in the example solution). If you compute x1-x2 = (3,3,6,6,-3), you find your vector multiplied by 3, so yes, your vector is also an s-invariant.
I don't see a simpler way than solving the linear equation system, albeit the cycles in the dataflow graph correspond with the s-invariants.
