LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

Methods for Constructing Coded DNA Languages 247
same probability 1
p . In this case the entropy essentially counts the average number
of words of a given length as subwords of the code words [17]. From the
coding theorem, it follows that {0, 1} ∗ can be encoded by X∗ with Σ ↦→ {0, 1} if
the entropy of X∗ is at least log 2 ([1], see also Theorem 5.2.5 in [6]). The codes
for θ-free, strictly θ-free, and θ-k-codes designed in this section have entropy
larger than log 2 when the alphabet has p = 4 symbols. Hence, such DNA codes
can be used for encoding bit-strings.
We start with the entropy definition as defined in [6].
Definition 31 Let X be a code. The entropy of X∗ is defined by
1
�(X) =limn→∞
n log |Subn(X ∗ )|.
If G is a deterministic automaton or an automaton with a delay that recognizes
X∗ and AG is the adjacency matrix of G, then by Perron-Frobenius theory
AG has a positive eigen value ¯µ and the entropy of X∗ is log ¯µ (see Chapter 4
of [6]). We will use this fact in the following computations of the entropies of
the designed codes. In [13], Proposition 16, authors designed a set of DNA code
words that is strictly θ-free. The following propositions show that in a similar
way we can construct codes with additional “good” propoerties.
In what follows we assume that Σ is a finite alphabet with |Σ| ≥3and
θ : Σ → Σ is an involution which is not identity. We denote with p the number
of symbols in Σ. WealsousethefactthatXis (strictly) θ-free iff X∗ is (strictly)
θ-free, (Proposition 4 in [14]).
Proposition 32 Let a, b ∈ Σ be such that for all c ∈ Σ \{a, b}, θ(c) /∈ {a, b}.
Let X = �∞ i=1 am (Σ \{a, b}) ibm for a fixed integer m ≥ 1.
Then X and X∗ are θ-free. The entropy of X∗ is such that log(p−2) < �(X ∗ ).
Proof. Let x1,x2,y ∈ X such that x1x2 = sθ(y)t for some s, t ∈ Σ + such
that x1 = ampbm , x2 = amqbm and y = amrbm ,forp, q, r ∈ (Σ \{a, b}) + .
Since θ is an involution, if θ(a) �= a, b, then there is a c ∈ Σ \{a, b} such
that θ(c) = a, which is excluded by assumption. Hence, either θ(a) = a or
θ(a) =b. Whenθis morphic θ(y) =θ(am )θ(r)θ(bm )andwhenθis antimorphic
θ(y) =θ(bm )θ(r)θ(am ). So, θ(y) =amθ(r)bm or θ(y) =bmθ(r)am .Sincex1x2 =
ampbmamqbm = sbmθ(r)am t or x1x2 = ampbmamqbm = samθ(r)bm t the only
possibilities for r are θ(r) =p or θ(r) =q. In the first case s =1andinthe
second case t = 1 which is a contradiction with the definition of θ-free. Hence X
is θ-free.
Let A = (V,E,λ) be the automaton that recognizes X∗ where V =
{1, ..., 2m +1} is the set of vertices, E ⊆ V × Σ × V and λ : E → Σ (with
(i, s, j) ↦→ s) is the labeling function.
An edge (i, s, j) isinEif and only if:
⎧
⎨ a, for 1 ≤ i ≤ m, j = i +1
s = b, for m +2≤ i ≤ 2m, j = i +1, and i =2m +1,j =1
⎩
s, for i = m +1,m+2,j= m +2,s∈ Σ \{a, b}