LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

282 Manfred Kudlek
Lemma 5. {[x]}◦{[y]} = κ(γ([x]) ([y])), A ◦ B = κ(γ(A) γ(B)), and{x1]}◦
···◦{[xk]} = κ(γ([x1]) ··· γ([xk])) for k ≥ 1.
Proof. The first statement is just the definition of ◦.
A ◦ B = �
�
[x]∈A,[y]∈B {[x]}◦{[y]} = [x]∈A,[y]∈B κ(γ([x]) γ([y]))
= κ( �
[x]∈A,[y]∈B γ([x]) γ([y])) = κ(γ(A) γ(B)).
The last statement follows by induction on k, using the second statement. ✷
3 Results
The first theorem presents a proof of Exercises 3.4 c in [6] or 4.2.11 in [5],
respectively. (We give a proof because of technical reasons.)
Theorem 1. REG is closed under cyclic permutation: γκ(REG) ⊂ REG.
Proof. Let G =(VN ,VT ,S,P) be a type-3 grammar generating the language
L = L(G). Assume that G is in normal form, i.e. the productions are of the form
A→aB or A→a, andS does not occur on the right-hand side of any production.
Assume also that each x ∈ VN ∪ VT is reachable from S.
From G construct a new type-3 grammar G1 =(V1N ,V1T ,S1,P1) with
V1N = VN × VN ∪ VN × VN ∪{S1},
V1T = VT , and productions
P1 = {S1→〈B,A〉 |∃A→aB ∈ P }
∪{〈B,A〉→a〈C, A〉 |B→aC ∈ P } ∪ {〈B,A〉→a〈 ¯ S,A〉 |B→a ∈ P }
∪{〈 ¯ B,A〉→a〈 ¯ C,A〉|B→aC ∈ P }∪ {〈 Ā, A〉→a | A→a ∈ P }.
Then a derivation
S⇒a0A1⇒···⇒a0 ···ak−1Ak⇒a0 ···akAk+1⇒···
⇒a0 ···akak+1 ···amAm+1⇒a0 ···akak+1 ···amam+1
implies a derivation
S1⇒〈Ak+1,Ak〉⇒ak+1〈Ak+2,Ak〉⇒ak+1 ···am〈Am+1,Ak〉
⇒ak+1 ···amam+1〈 ¯ S,Ak〉⇒ak+1 ···am+1a0〈 Ā1,Ak〉
⇒···ak+1 ···am+1a0 ···ak−1〈 Āk,Ak〉⇒ak+1 ···am+1a0 ···ak,
and therefore
S1⇒∗ ak+1 ···am+1a0 ···am.
On the other hand, a derivation
S1⇒〈Ak+1,Ak〉⇒ak+1〈Ak+2,Ak〉⇒ak+1 ···am〈Am+1,Ak〉
⇒ak+1 ···amam+1〈 ¯ S,Ak〉⇒ak+1 ···am+1a0〈 Ā1,Ak〉
⇒···ak+1 ···am+1a0 ···ak−1〈 Āk,Ak〉⇒ak+1 ···am+1a0 ···ak
implies derivations
Ak+1⇒···⇒ak+1 ···amAm+1⇒ak+1 ···amam+1 and
S⇒a0A1⇒···a0 ···ak−1Ak⇒a0 ···ak−1akAk+1,
and therefore
S⇒∗ a0 ···akak+1 ···am+1.
Thus, L(G1) =γκ(L), the set of all cyclic permutations of L.