LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

On Some Classes of Splicing Languages 89
splicing rules are one-sided splicing rules of radius k = 1. Moreover, the “onesidedness”
of the rules considered by Goode and Pixton is of the (1, 3) type. We
present the basic definitions for the four possible types.
For two symbols a, b ∈ V a semi-simple (linear) splicing rule with markers a
and b is a splicing rule of one of the following four types: (a, 1; b, 1), (1,a;1,b),
(1,a; b, 1), (a, 1; 1,b).
The action of the four rules on strings is obvious.
A semi-simple H system is an H system S =(V,A,R) with all rules in R
semi-simple splicing rules of one of the types (i, j), with i =1, 2, j =3, 4.
The language generated by a semi-simple H system S as above is defined as
usual for splicing schemes, L = σ ∗ (A), where σ =(V,R). If the rules in R are
semi-simple rules of type (i, j), this language will be called an (i, j)-semi-simple
splicing language. Theclassofall(i, j)-semi-simple splicing languages will be
denoted by SSH(i, j).
In [8] only the class SSH(1, 3) is considered. Since in [8] only one example
of a semi-simple language of the (1, 3) type is given, we find it useful to provide
some more.
Example 1 Consider S3 =({a, b}, {aba}, {(a, 1; b, 1)}). The language generated
by S3 is L3 = L(S3) =aa + ∪ aba + ∈ SSH(1, 3).
Example 2 Consider S4 =({a, b}, {aba}, {(1,a;1,b)}). The language generated
by S4 is L4 = L(S4) =b + a ∪ ab + a ∈ SSH(2, 4).
The following result shows that, even for the types (1, 3) and (2, 4), which
share some similarity of behavior, unlike in the case of simple rules, the respective
classes of languages are incomparable.
Theorem 2. The classes SSH(1, 3) and SSH(2, 4) are incomparable.
Proof: Note first that simple splicing languages in the class SH are in the intersection
SSH(1, 3) ∩ SSH(2, 4).
The language L1 = a + ∪ a + ab ∪ aba + ∪ aba + b belongs to SSH(1, 3), but not
to SSH(2, 4).
For the first assertion, L1 is generated by the (1, 3)-semi-simple H system
S1 =({a, b}, {abab}, {(a, 1; b, 1)}). We sketch the proof of L1 = L(S1). Denote
by r the unique splicing rule, r =(a, 1; b, 1). For the inclusion L1 ⊆ L(S1), note
first that
(a|bab, abab|) ⊢r a,
(a|bab, ab|ab) ⊢r a 2 b.
Next, if a n b ∈ L(S1), then a n ,a n+1 b ∈ L(S1), for any natural n ≥ 2, since
(a n |b, abab|) ⊢r a n ,
(a n |b, ab|ab) ⊢r a n+1 b.
Thus, by an induction argument, it follows that a + ⊆ L(S1) anda + ab ⊆ L(S1).