LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

124 Karel Culik II, Juhani Karhumäki, and Petri Salmela
u3 = ab : a · ab · (bab) n /∈ Suf(X + ) implies aab(bab) n /∈ C(X) sothat
aab(bab) n /∈ XC(X) and therefore ab(bab) n /∈ C(X) for all n ∈ N.
Hence the number n3 does not exist.
u4 = ba : ba · an · a/∈ XC(X) and therefore ba · an /∈ C(X) for all n ∈ N,
and hence the number n4 does not exist.
u5 = bb : bb ∈ X ⊆ C(X) sothatn5 =0.
As a conclusion I = {0, 1, 5}, andG = �
i∈I uiX ni =1· X + a + bb = X. This
gives us the centralizer
in other words we have established:
C(X) =GX ∗ = XX ∗ = X + ,
Fact 1 C({a, bb, aba, bab, bbb})={a, bb, aba, bab, bbb} + .
Next we prove that the fixed point approach applied to the language X leads
to an infinite loop of iterations. We prove this by showing that there exist words
in C(X) \ Xi for every Xi of the iteration (1). To do this we take a closer look on
the language L =(bab) ∗ ab(bab) ∗ . Clearly L ⊆ X0 =Pref(X + ) ∩ Suf(X + )and
L ∩ X + = ∅.
By the definition of the fixed point approach, word w ∈ Xi is in Xi+1 if and
only if Xw ⊆ XiX and wX ⊆ XXi. We will check this condition for an arbitrary
word (bab) k ab(bab) n ∈ L with k, n ≥ 1. The first condition Xw ⊆ XiX leads to
the cases:
and
a · (bab) k ab(bab) n =(aba)(bb · a) k−1 (bab) n+1 ∈ X + X ⊆ XiX,
bb · (bab) k ab(bab) n =(bbb)a(bb · a) k−1 (bab) n+1 ∈ X + X ⊆ XiX,
aba · (bab) k ab(bab) n = a(bab)a(bb · a) k−1 (bab) n+1 ∈ X + X ⊆ XiX,
bbb · (bab) k ab(bab) n =(bb) 2 a(bb · a) k−1 (bab) n+1 ∈ X + X ⊆ XiX
bab · (bab) k ab(bab) n =(bab) k+1 ab(bab) n−1 · bab ∈ XiX.
However, the last one holds if and only if
and
⎫
⎪⎬
⎪⎭
(3)
(bab) k+1 ab(bab) n−1 ∈ Xi. (4)
Similarly, the second condition wX ⊆ XXi yields us:
(bab) k ab(bab) n · a = bab(bab) k−1 (a·bb) n aba ∈ XX + ⊆ XXi,
(bab) k ab(bab) n · bb = bab(bab) k−1 (a·bb) n a · bbb ∈ XX + ⊆ XXi,
(bab) k ab(bab) n · aba = bab(bab) k−1 (a·bb) n a·a·bab·a ∈ XX + ⊆ XXi,
(bab) k ab(bab) n · bbb = bab(bab) k−1 (a·bb) n a·a·bb· bb ∈ XX + ⊆ XXi
(bab) k ab(bab) n · bab = bab · (bab) k−1 ab(bab) n+1 ∈ XXi.
⎫
⎪⎬
⎪⎭
(5)