LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

Communicating Distributed H Systems with Alternating Filters 381
Molecules remaining in tube 1:
R ′ S ′ Z1,R ′ 1 Sqjw2Y,Xw1alSqjw2Y.
Molecules going from tube 2 to tube 1:
RL.
Molecules remaining in tube 2:
R ′ L ′ ,R R 1 alS ′ qjw2Y,Xw1alS ′ qjw2Y,Xw1alL ′ ,R ′ S ′ qjw2Y,R ′ 1 L′ ,
Xw1Sqiakw2Y,Xw1L, RSqiakw2Y,R R 1 L.
On the next iteration R ′ 1Sqjw2Y will go from the first test tube to the second
tube.
Thus we modelled the application of the rule (qi,ak,R,al,qj) of the Turing
machine M. It is easy to observe that if w2 = ε we first apply rule 1.1.1.1 in
order to extend the tape to the right and after that w2 = a0. Alsoitiseasy
to check that additional molecules obtained (such as Xw1Sqiakw2Y, Xw1L,
RSqiakw2Y, R R 1 L, R R 1 alS ′ qjw2Y,RSqiakZ R 1 ,Xw1alS ′ qjw2Y, R ′ S ′ qjw2Y,
R ′ S ′ Z1, R ′ 1 Sqjw2Y, Xw1alS ′ qjw2Y,Xw1alL ′ ) either go to the second test tube
or do not alter the computation.
For the left shift rules the things are similar, except that we ensure that there
are at least 2 symbols at the left of S.
We see that we model step by step the application of rules of M. Itiseasy
to observe that this is the only possibility for the computations to go on as the
molecule which encodes the configuration of M triggers the application of rules
of Γ .
Theorem 4. For any recursively enumerable language L⊆T ∗ there is a communicating
distributed H system with alternating filters with two components and
two filters Γ =(V,T,(A1,R1,F (1)
1 ,F (2)
1 ), (A2,R2,F (1)
2 ,F (2)
2 )), which generates
L, i.e., L = L(Γ ).
Proof. For the proof we use ideas similar to ones used in [10].
Let L = {w0,w1,...}. Then there is a Turing machine TL which computes
wi0 ...0startingfrom01i+1 where 0 is the blank symbol.
It is possible to construct such a machine in the following way. Let G be a
grammar which generates L. Then we can construct a Turing machine M which
will simulate derivations in G, i.e., we shall obtain all sentential forms of G. We
shall simulate bounded derivation in order to deal with a possible recursion in
a derivation. After deriving a new word the machine checks if this word is a
terminal word and, if this is the case, it checks if this word is the i-th terminal
word which is obtained. If the last condition holds, then the machine erases
everything except the word.
Moreover, this machine is not stationary (i.e., the head has to move at each
step) and it never visits cells to the left of the 0 from the initial configuration
(i.e., the tape is semi-infinite to the right). Ideas for doing this can be found
in [8]. So, machine TL transforms configuration q001k+1 into qf wk0 ...0.
Now starting from TL it is possible to construct a machine T ′ L which computes
01 k+2 Mq ′ f wk0 ...0 starting from q001 k+1 .Now,usingΓ we shall cut off wk and
pass to configuration q001 k+2 . In this way we will generate L word by word.