If two transitions t1, t2 have the same action and the same exit condition, then (maybe) we could remove t2 and update all transitions leading to t2 so that their exit condition leads to t1.
If we remove t_b2, and fix the exit condition of t_c, we get:
<1, a, 2 \/ 3>
<2, b, 0>
<3, c, 2>
which still captures the behaviour of the process.
Rationale
This reduction could be applied repeatedly until no more triples are removed. If two agents share (partially or completely) the same behaviour, the reduction should significantly reduce the number of triples needed to encode the whole system
If two transitions
t1,t2have the same action and the same exit condition, then (maybe) we could removet2and update all transitions leading tot2so that their exit condition leads tot1.A simple case
Consider
a.(b.0 + c.b.0). This gets encoded asIf we remove
t_b2, and fix the exit condition oft_c, we get:which still captures the behaviour of the process.
Rationale
This reduction could be applied repeatedly until no more triples are removed. If two agents share (partially or completely) the same behaviour, the reduction should significantly reduce the number of triples needed to encode the whole system