Trivial cycle resolution

[mscuttari]

Considering the system of equations:

x + y + z = 1 // eq1: matched x
x + y - z = 2 // eq2: matched y
x - y + z = 3 // eq3: matched z

Explicitating eq1 with respect to x would give x = - y - z + 1
But the substitution within eq2 would yield to - y - z + 1 + y - z = 2
Which is equivalent to 0y - 2z + 1 = 0
And would lead to the division by 0 y = (2z - 1) / 0

The system, however, is trivially solvable with x = 5/2, y = -1, z = -1/2

[mscuttari]

Commit 7df8123 introduces the detection of division by zero. This allows to stop the compilation process if the issue is detected.

An implementation of the Cramer's rule can now be introduced to solve this issue without relying to external solvers.

[casella]

@mscuttari is this issue still open?

If so, do we wait for @StefanoAzzone to finish his job in order to have a proper solution?

[mscuttari]

Yes, the solution is part of Stefano's work.