LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

214 Masami Ito and Ryo Sugiura
Lemma 21 Let A, B ⊆ X ∗ be regular languages. Then A ⋄ B is a regular language.
Proof. By X we denote the new alphabet {a | a ∈ X}. LetA =(S, X, δ, s0,F)
be a finite deterministic automaton with L(A) =A and let B =(T,X,θ,t0,G)
be a finite deterministic automaton with L(B) =B. Define the automaton B =
(T,X,θ, t0,G) as θ(t, a) = θ(t, a) for any t ∈ T and a ∈ X. Letρ be the
homomorphism of (X ∪ X) ∗ onto X ∗ defined as ρ(a) =ρ(a) =a for any a ∈ X.
Moreover, let L(B) =B. Thenρ(B) ={ρ(u) | u ∈ B} = B and ρ(A⋄B) =A⋄B.
Hence, to prove the lemma, it is enough to show that A ⋄ B is a regular language
over X ∪ X. Consider the automaton A⋄B =(S × T,X∪ X,δ⋄ θ, (s0,t0),F× G)
where δ ⋄ θ((s, t),a)=(δ(s, a),t)andδ ⋄ θ((s, t), a) =(s, θ(t, a)) for any (s, t) ∈
S×T and a ∈ X. Then it is easy to see that w ∈L(A⋄B) if and only if w ∈ A⋄B,
i.e., A ⋄ B is regular. This completes the proof of the lemma.
Proposition 21 Let A, B ⊆ X ∗ be regular languages and let n be a positive
integer. Then A ✄ [n] B is a regular language.
Proof. Let the notations of X, B and ρ be the same as above. Notice that
A ✄ [n] B =(A ⋄ B) ∩ (X ∗ X ∗ ) n X ∗ .Since(X ∗ X ∗ ) n X ∗ is regular, A ✄ [n] B is
regular. Consequently, A ✄ [n] B = ρ(A ✄ [n] B)isregular.
Remark 21 The n-insertion of a context-free language into a context-free language
is not always context-free. For instance, it is well known that A = {a n b n |
n ≥ 1} and B = {c n d n | n ≥ 1} are context-free languages over {a, b} and {c, d},
respectively. Since (A ✄ [2] B) ∩ a + c + b + d + = {a n c m b n d m | n, m ≥ 1} is not
context-free, A✄ [2] B is not context-free. Therefore, for any n ≥ 2, n-insertion of
a context-free language into a context-free language is not always context-free.
However, A ✄ [1] B is a context-free language for any context-free languages A
and B (see [4]). Usually, a 1-insertion is called an insertion.
Now consider the n-insertion of a regular (context-free) language into a
context-free (regular) language.
Lemma 22 Let A ⊆ X ∗ be a regular language and let B ⊆ X ∗ be a context-free
language. Then A ⋄ B is a context-free language.
Proof. The notations which we will use for the proof are assumed to be the
same as above. Let A =(S, X, δ, s0,F) be a finite deterministic automaton with
L(A) =A and let B =(T,X,Γ,θ,t0,ɛ) be a pushdown automaton with N (B) =
B. LetB =(T,X,Γ,θ, t0,γ0,ɛ) be a pushdown automaton such that θ(t, a, γ) =
θ(t, a, γ) for any t ∈ T,a ∈ X ∪{ɛ} and γ ∈ Γ .Thenρ(N (B)) = B. Now define
the pushdown automaton A⋄B =(S × T,X ∪ X,Γ ∪{#},δ⋄ θ, (s0,t0),γ0,ɛ)as
follows:
1. ∀a ∈ X, δ ⋄ θ((s0,t0),a,γ0) ={((δ(s0,a),t0), #γ0)},
δ ⋄ θ((s0,t0), a, γ0) ={((s0,t ′ ), #γ ′ ) | (t ′ ,γ ′ ) ∈ θ(t0, a, γ0)}.
2. ∀a ∈ X, ∀(s, t) ∈ S × T,∀γ ∈ Γ ∪{#},δ⋄ θ((s, t),a,γ)={((δ(s, a),t),γ)}.