LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

where
Solving Graph Problems by P Systems 15
O = {ai,di,ti,t ′ i ,fi | 1 ≤ i ≤ n}∪{hi,h ′ i | 0 ≤ i ≤ k}∪{ei | 0 ≤ i ≤ m}
∪{gi | 1 ≤ i ≤ m +1}∪{g ′ i | 1 ≤ i ≤ m}∪{a, b, p, q, u, yes},
H = {1, 2},
µ =[ 1 [ 2 ] 0 0
2 ] 1 ,
w1 = λ,
w2 = a1a2 ···and1,
and the set R contains the following rules
1. [ 2ai] 0
2 → [ 0
2ti] 2 [ 0
2fi] 2 , 1 ≤ i ≤ n.
The objects ai correspond to vertices vi, 1 ≤ i ≤ n. By using a rule as
above, for i non-deterministically chosen, we produce the two objects ti and
fi associated to vertex vi, placed in two separate copies of membrane 2,
where ti means that vertex vi appears in some subset of vertices, and fi
means that vertex vi does not appear in some subset of vertices. Note that
the charge remains the same for both membranes, namely neutral, hence
the process can continue. In this way, in n steps we get all 2n subsets of
V , placed in 2n separate copies of membrane 2. In turn, these 2n copies
of membrane 2 are within membrane 1 – the system always has only two
levels of membranes. Note that in spite of the fact that in each step the
object ai is non-deterministically chosen, after n steps we get the same result,
irrespective of which objects were used in each step.
2. [ 2di → di+1] 0
2 , 1 ≤ i ≤ n − 1.
3. [ 2dn → qqh0] 0
2 .
The objects di are counters. Initially, d1 is placed in membrane 2. By division
(when using rules of type 1), we introduce copies of the counter in each new
membrane. In each step, in each membrane with label 2, we pass from di to
di+1, thus “counting to n”. In step n we introduce both q and h0; the objects
q will exit the membrane (at steps n +1,n + 2), changing its polarization,
the object h0 is a new counter which will be used at the subsequent steps as
shown below. Note that at step n we have also completed the generation of
all subsets.
Now we pass to the next phase of computation – counting the number of
objects ti (1 ≤ i ≤ n) in each membrane with label 2, which corresponds to
the cardinality of each subset; we will select out the subsets with cardinality
exactly k.
4. [ 2q] 0
2 → [ −
2 ] 2 u.
5. [ 2q] −
2 → [ 0
2 ] 2u. 6. [ ti → abt 2 ′ −
i ] , 1 ≤ i ≤ n.
2
7. [ 2hi → h ′ 0
i ] 2 , 0 ≤ i ≤ k.
8. [ 2h ′ +
i → hi+1]
2 , 0 ≤ i ≤ k − 1.
9. [ 2a] 0
2 → [ +
2 ] 2 u.