LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

On Languages of Cyclic Words 283
The proper inclusion follows from the fact that there are regular languages
not being closed under cyclic permutation, like L = {a m b n | m, n ≥ 0}. ✷
Theorem 2. LIN is not closed under cyclic permutation, i.e., γκ(LIN) �⊂
LIN.
Proof. Consider the language L = {a n b n | n ≥ 0} ∈LIN. Assume γκ(L) ∈
LIN. Thus,a k b 2k a k ∈ [a 2k b 2k ]. By the iteration lemma for linear context-free
languages there exists N = N(L) ∈ IN such that for k ≥ N there are words
u, v, w, x, y such that z = a k b 2k a k = uvwxy with |uv| ≤N, |xy| ≤N, |vx| > 0,
and ∀j ≥ 0:uv j wx j y ∈ L. Butthenuv ∈{a} ∗ , xy ∈{a} ∗ ,andonlya’s can be
iterated, contradicting the fact that |z|a = |z|b for all z ∈ γκ(L). ✷
The next Theorem is from [6].
Theorem 3. CF is closed under cyclic permutation: γκ(CF) ⊂ CF.
Proof. γκ(CF) ⊂ CF is Exercise 6.4 c and its solution in [6], pp. 142-144. The
problem is also stated as Exercise 7.3.1 c in [5].
The proper inclusion follows again from the fact that there are context-free
(even regular) languages not being closed under cyclic permutation, like the
example in Theorem 3.1. ✷
Theorem 4. κREG �⊂ RAT(◦).
Proof. Consider L = {(ab) n | n ≥ 0} ∈REG. From this follows that κ(L) =
{[(ab) n ] | n ≥ 0} with [(ab) n ]={(ab) n , (ba) n }.
Now assume κ(L) ∈ RAT(◦). Then, by the iteration lemma for RAT(◦)
there exists a N ∈ IN such that for k>Nthere exist u, v, w such that [(ab) k ] ∈
{[u]} ◦{[v]} ◦{[w]} with ||{[u]} ◦{[v]}|| ≤ N, ||[v]|| > 0, and with ∀j ≥ 0:
{[u]}◦{[v]} j ◦{[w]} ⊆κ(L).
Now [uvw] ∈ κ(L) and[uvvw] ∈ κ(L). Because of ||[v]|| > 0 it follows that
v = av ′ or v = bv ′ .From[v ′ a]=[av ′ ]and[v ′ b]=[bv ′ ] follows that [uv ′ aav ′ w] ∈
κ(L) or[uv ′ bbv ′ w] ∈ κ(L), a contradiction to the structure of κ(L). ✷
Theorem 5. γRAT(◦) �⊂ CF.
Proof. Consider the language M of cyclic words defined by M = {[a]}◦M ◦{[bc]}.
Clearly,
M = �
k≥0
{[a]} k ◦{[bc]} k .
Now, {[a]} k = {[a k ]}, and[b k c k ] ∈{[bc]} k for k ≥ 0 which can be shown
by induction on k, namely[bc] ∈{[bc]}, and assuming [b k c k ] ∈{[bc]} k follows
[b k+1 c k+1 ]=[b k c k cb] ∈{[b k c k ]}◦{[bc]} ⊆{[bc]} k+1 .
From this follows that [a k b k c k ] ∈ {[a k ]}◦{[b k c k ]} = {[a]} k ◦{[b k c k ]} ⊆
{[a]} k ◦{[bc]} k ⊆ M.
Therefore, γ(M) ∩{a} ∗ {b} ∗ {c} ∗ = {a k b k c k | k ≥ 0} �∈ CF.
Thus, γ(M) �∈ CF. ✷