LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

16 Artiom Alhazov, Carlos Martín-Vide, and Linqiang Pan
10. [ 2b] +
2 → [ 0
2 ] 2u. At step n +1, h0 evolves to h ′ 0 , and at the same time one copy of q exits the
membrane, changing its polarization to negative. At step n +2, ti evolves to
abt ′ i , and at the same time the other copy of q exits the membrane, changing
its polarization to neutral. At step n +3, onecopyofaexits the membrane,
changing its polarization to positive. At step n +4,h ′ 0 evolves to h1, andat
thesametimeonecopyofbexits the membrane, returning its polarization
to neutral (this makes possible the use of rules of types 7 and 9).
The rules of types 7, 8, 9, and 10 are applied as many times as possible (in
one step rules of types 7 and 9, in the next one rules of types 8 and 10, and
then we repeat the cycle). Clearly, at step n +2+2k, a membrane contains
object hk if and only if the cardinality of the corresponding subset is at
least k. Atstepn +3+2k, in the membranes whose corresponding subsets
have cardinality more than k, hk evolves to h ′ k , and one copy of a changes
their polarization to positive. These membranes will no longer evolve, as no
further rule can be applied to them. In the membranes whose corresponding
subsets have cardinality exactly k, hk evolves to h ′ k , and their polarization
remains neutral, because there is no copy of a which can be used. We will
begin the next phase of computation with the rule of type 11 – checking
whether a subset with cardinality k is a vertex cover.
11. [ 2h ′ 0
k → qqg1] 2 .
12. [ 2t ′ i → ei1
−
···eil ] 2 ,1≤ i ≤ n, andvertexviis adjacent to edges ei1 ···eil .
At step n +4+2k, in the membranes with label 2 and polarization 0, h ′ k
evolves to qqg1.Atstepn+5+2k,onecopyofqexits the membrane, changing
its polarization to negative. At step n +6+2k, in parallel t ′ i (1 ≤ i ≤ n)
evolves to ei1 ···eil , and at same time the other copy of q exits the membrane,
changing its polarization to neutral. After completing this step, if there is at
least one membrane with label 2 which contains all symbols e1, ··· ,em, then
this means that the subset corresponding to that membrane is a vertex cover
of cardinality k. Otherwise (if in no membrane we get all objects e1, ··· ,em),
there exists no vertex cover of cardinality k. In the following steps, we will
“read” the answer, and send a suitable signal out of the system. This will be
done by the following rules.
13. [ 2gi → g ′ 0
ip] 2 , 1 ≤ i ≤ m.
14. [ 2e1] 0
2 → [ +
2 ] 2 u.
Object gi evolves to g ′ ip. At the same time for all subsets of cardinality k, we
check whether or not e1 is present in each corresponding membrane. If this is
the case, then one copy of e1 exits the membrane where it is present, evolving
to u, and changing in this way the polarization of that membrane to positive
(the other copies of e1 will immediately evolve to e0, which will never evolve
again). The membranes which do not contain the object e1 remain neutrally
charged and they will no longer evolve, as no further rule can be applied to
them.
15. [ 2ei → ei−1] +
2 , 1 ≤ i ≤ m.
16. [ 2g ′ i , 1 ≤ i ≤ m.
→ gi+1] +
2