13th International Conference on Membrane Computing - MTA Sztaki

D. Sburlan
In case of M, the states and the transitions between them are defined as
follows.
{X 1−, X 2−, X 3−, l 2 ↑}, op
where op = inc if r = 1
and op = skip if r ≠ 1
l 2
{l 1 ↓, l 1 ↑, a 1−, a 2−, a 3−}, skip
l 1 l 1
{X 1−, X 2−, X 3−, l 3 ↑}, op
where op = inc if r = 1
and op = skip if r ≠ 1
l 3
• for the instruction l h : halt ∈ P
l h → λ
a i → a i , 1 ≤ i ≤ 3
In case of M, the states and the transitions between them are defined as
follows.
l h
{l h ↓, a 1−, a 2−, a 3−}, skip
l H
Here is shown how Φ(Π, M) works. At the beginning of a computation in
the region 1 of Π there exists the multiset composed by just one object l 0 (that
corresponds to the label of the first register machine instruction). This object
will be iteratively rewritten during the computation (according with the register
machine program) into the label of an instruction. In any configuration in a
computation of Π, the number of objects a r corresponds to the number stored
in register r, 1 ≤ r ≤ 3. Following the register machine definition, in the initial
configuration there will be no objects a r , 1 ≤ r ≤ 3, because the register machine
M starts with all registers being empty.
Assume now that the current register machine instruction to be simulated is
l 1 : (add(r), l 2 , l 3 ); then Π is in a configuration C = l 1 a k1
1 ak2 2 ak3 3 and M is in the
state labeled l 1 . In this configuration, Π executes the rules l 1 → l 1 and a r → a r ,
for 1 ≤ r ≤ 3. Consequently, the next configuration is C ′ = l 1 a k1
1 ak2 2 ak3 3 , hence
M passes from state l 1 to state l 1 . Next, Π non-deterministically executes one
of the rules l 1 → l 2 a r and l 1 → l 3 a r (exactly one of them, because there is
only one object l 1 ), and the rules a r → a r ; in this way the next configuration
will be C ′′ = l 2 a k1
1 ak2 2 ak3 3 a r or C ′′ = l 3 a k1
1 ak2 2 ak3 3 a r where 1 ≤ r ≤ 3. It follows
that M passes from state l 1 to the state l 2 or l 3 , therefore the simulation of the
addition instruction was correctly performed. However, as we will see later on,
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