LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

Communicating Distributed H Systems with Alternating Filters 375
w will be a result. So, now we shall show how to do this. Suppose that XwBY
appears for the first time in tube 1 at some step q (it is easy to check that q is
even).
Step q.
Tube 1.
XwBY h −→
1.3 Xw ⋄ Yβh −→
1.5 Xβββw ⋄ Yβ2.
We can also have:
Xw ⋄ Yβh −→
1.4 Xαβw ⋄ Yβ ↑.
Xβββw ⋄ Yβ can pass filter F (2)
2 .
Step q+1.
Tube II.
Xβββw ⋄ Yβh −→
2.5 w ⋄ Yβh −→
2.6 w1.
So, if w ∈ T ∗ then w is a result.
Final Remarks We start with molecule XBSY in the first tube. After that
we are doing a rotation and simulation of productions of G. Finally we reach a
configuration XwBY where w ∈ T ∗ . At this moment we eliminate X, Y and B
obtaining w ∈ T ∗ .
As all cases have been investigated, the proof is complete.
4 TTF2,4 with Garbage Collection
In this section we shall present a TTF system having two components and four
filters. Additionally this system contains no rules in the second component, so
it is used only as a garbage collector. Also, all filters in the second tube are the
same.
Theorem 2. For any type-0 grammar G =(N,T,P,S) there is a communicating
distributed H system with alternating filters having two components and four
filters Γ =(V,T,(A1,R1,F (1)
1 ,...,F (4)
1 ), (A2,R2,F (1)
2 ,...,F (4)
2 )) which simulates
G and L(Γ ) = L(G). Additionally, the second component of Γ has no
splicing rules, i.e., R2 = ∅, andalsoF (1)
(4)
2 = ···= F 2 .
We construct Γ as follows.
Let N ∪ T ∪{B} = {a1,a2,...,an} (B = an).
In what follows we will assume the following:
1 ≤ i ≤ n, 1 ≤ j ≤ 4, a ∈ N ∪ T ∪{B},
b ∈ N ∪ T ∪{B} ∪ {⋄},γ ∈{α, β}.
Also let V = N ∪ T ∪{B}∪{α, β} ∪ {⋄}.
V = V∪{X, Y, Xα,Xβ,X ′ α ,Yα,Y ′ α ,Yβ,Z,Z ′ ,Zj,Rj,Lj}.
The terminal alphabet T is the same as for grammar G.