13th International Conference on Membrane Computing - MTA Sztaki

R. Pagliarini, O. Agrigoroaiei, G. Ciobanu, V. Manca
The underlining idea for this definition is that when some (multiset of) objects
v appear during the course of an evolution of a P system, we look for some
(multiset of) rules G which have produced them. By producing we understand
that we have an evolution step u =⇒ F
u ′ in which u ′ ≥ v and F ≥ G such
that exactly the rules in G are those responsible for the apparition of v. Note
that v can be written as the sum of v ∩ rhs(G) and v\rhs(G). The v ∩ rhs(G)
part is the one produced by G since it is included in rhs(G); for the remainder
v\rhs(G) we require that it is composed only of objects which do not evolve in
the considered evolution step (this is what the first condition amounts to). In
other words, all the objects of v are either produced by rules of G or are not
interacting with any rule. The second condition is equivalent to saying that no
rule r can be subtracted from G such that the part of v produced by G remains
the same - there are no “useless” rules in G with respect to producing elements
of v.
To view the notions above introduced, let us consider the following example
of a transition P system with only one membrane, with rules
r 1 : x → a + b, r 2 : y → b, r 3 : a + b → y
is
and an initial multiset of objects u 0 = x+y +2a. The only possible evolution
x + y + 2a r1+r2
=⇒ 3a + 2b =⇒ 2r3
a + 2y =⇒ 2r2
a + 2b =⇒ r3
b + y =⇒ r2
2b
In [2], a general inductive procedure for finding the causes of a multiset has
been introduced. Here we reason directly over the example considered, loosely
following the inductive procedure.
Let v = a + b be the multiset for which we intend to find its causes. We
start by considering the empty multiset 0 as a potential cause for v. The empty
multiset is discarded because lhs(r 3 ) ≤ v\rhs(0) = v\0 (they are actually equal)
which contradicts the first part of Definition 1. The next possible candidates for
causes of v are either r 1 or r 2 or r 3 . Clearly r 3 suffers from the same problem as
0, it does not fulfill the first condition of Definition 1. However, both r 1 and r 2
verify the conditions to be causes of v. The next step is to add rule r 3 to either
multiset r 1 or r 2 , i.e., we consider as potential causes r 1 +r 3 and r 2 +r 3 . We find
that rule r 3 is actually “useless” as it does not produce any object of v. In other
words, r 3 has the problem that rhs(G) ∩ v = rhs(G − r 3 ) ∩ v for G = r 1 + r 3
or G = r 2 + r 3 . Moreover, this happens for any G ≥ r 1 + r 3 or G ≥ r 2 + r 3 .
This means that no cause of v can contain r 3 . If we try G = r 1 + r 2 , then r 2
becomes the “useless” rule: rhs(G) ∩ v = rhs(G − r 2 ) ∩ v. This also happens
for G ≥ r 1 + r 2 . All we have left to check are either G = k · r 1 or G = k · r 2 ,
for k ≥ 2. When G = k · r 1 we have that an instance of r 1 is a “useless” rule:
rhs(G)∩v = rhs(G−r 1 )∩v; in other words, any additional r 1 besides a single r 1
are “useless”. The case of G = k · r 2 for k ≥ 2 is similar. Thus the only possible
causes for v = a + b are either r 1 or r 2 .
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