LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

374 Sergey Verlan
Step 2.
Tube II.
Xαβwβα i−1 Y ′ αh −→
2.1 Xαβwβα i−1 Yαh −→
2.2 X ′ ααβwβα i−1 Yα1.
Only molecule X ′ α αβwβαi−1 Yα can pass filter F1.
In this way we transferred one α from the right end to the left end.
Step 3.
Tube I.
X ′ α αβwβαi−1 Yαh −→
1.9 X ′ α αβwβαi−2 Y ′ α
h −→
1.7 Xααβwβα i−2 Y ′ α 2.
We can also have:
X ′ ααβwβαi−1 Yαh −→ Xββαβwβα
1.8 i−1Yαh −→ Xββαβwβα
1.9 i−2Y ′ α ↑.
Only Xααβwβαi−1 Y ′ α
can pass filter F (1)
2 .
Step 4.
Tube II.
Xααβwβα i−2 Y ′ αh −→
2.1 Xααβwβα i−2 Yαh −→
2.2 X ′ αααβwβα i−2 Yα1.
Only molecule X ′ αααβwβα i−2 Yα can pass filter F1.
In this way we transferred one more α from the right end to the left end. We
can continue in this way until we obtain for the first time X ′ α αi βwβYα in tube 2
at some step p. This molecule is communicated to tube 1 where we can use the
rule 1.6 and we complete the rotation of ai.
Step p+1.
Tube I.
X ′ ααiβwβYαh −→ X
1.6 ′ ααiβwYβh −→ Xββα
1.8 iβwYβ2. We can also have:
X ′ ααiβwβYαh −→ Xαα
1.7 iβwβY ′ αh −→ Xαα
1.6 iβwYβ ↑.
Only XββαiβwYβ can pass filter F (2)
2 .Wealsoaddthatp+1 is an odd number,
so molecule XββαiβwYβ will remain for one step in the first component
because the filter used in the second component during odd steps is F (1)
2 .During
next step (p+2) this molecule will not change and it will be communicated
to the second tube because the filter to be used will be F (2)
2 .
Step p+3.
Tube II.
XββαiβwYβh −→ XaiwYβh −→ XaiwY1.
2.4 2.3
XaiwY goes to the first tube.
In this way we competed the rotation of ai.
Simulation of Grammar Productions If we have a molecule XwuY in the
first tube and there is a rule u → v ∈ P , then we can apply rule 1.1 to simulate
the corresponding production of the grammar.
Obtaining Results Now we shall show how we can obtain a result. If we have
a molecule of type XwBY , then we can perform a rotation of B as described
above. We can also take off X and Y from the ends. In this way, if w ∈ T ∗ ,then