LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

Fixed Point Approach to Commutation of Languages 123
Proof. If w ∈ C(X) andvis an incomparable element in X, thenequation
XC(X) =C(X)X implies that vw ∈ C(X)X and therefore vwv −1
1 ∈ C(X) for
some element v1 ∈ X. Repeating the argument n times we obtain
v n w(vn ···v2v1) −1 ∈ C(X), vi ∈ X,
where t = vn ···v2v1 and w = ut. Thenv n u ∈ C(X) for some integer n ∈
{0, 1, 2,...} and word u ∈ Suf(X)∩Pref(w). Since v is incomparable, we conclude
that for every s ∈ X n
and hence
v n us ∈ C(X)X n = X n C(X),
us ∈ C(X).
In other words, uX n ⊆ C(X). Since C(X) is a semigroup, we have also the
inclusion uX n X ∗ ⊆ C(X).
For every proper suffix ui ∈ Suf(X), including the empty word 1, there
either exists a minimal integer ni, forwhichuiX ni ⊆ C(X), or uiX n �⊆ C(X) for
every integer n ≥ 0. Since Lemma 2 excludes the latter case, we can associate
with every word w ∈ C(X) awordui ∈ Suf(X) and the minimal ni such that
w ∈ uiX ni X ∗ .
Lemma 3. If the finite language X contains an incomparable word, it has a
rational centralizer. Moreover, the centralizer is finitely generated.
Proof. If the language X is finite, then the set of proper suffixes of X is also
finite. With the above terminology we can write
C(X) = �
uiX
i∈I
ni X ∗ =( �
uiX
i∈I
ni ) X
� ��
=G
�
∗ = GX ∗ ,
where I is an index set defining suffixes ui above. Here the language G is finite
and X ⊆ G. Indeed if u0 =1,thenn0 = 1, and hence u0X n0 =1· X = X ⊆ G.
Since C(X) issemigroupandX is included in G, weobtain
C(X) =C(X) + =(GX ∗ ) + =(X + G) + = G + .
Now we can prove that the centralizer of our language X = {a, bb, aba, bab,
bbb} is X + .Thewordbab is incomparable. The set of proper suffixes of X is
{1,a,b,ab,ba,bb}. We will consider all of these words separately:
u0 =1: 1· X ⊆ C(X) sothatn0 =1.
u1 = a : a ∈ X ⊆ C(X) sothatn1 =0.
u2 = b : b · a n · a/∈ XC(X) =C(X)X and therefore b · a n /∈ C(X) for all n ∈ N.
This means that the number n2 does not exist.