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LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

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Fixed Point Approach to Commutation <strong>of</strong> Languages 121<br />

Here Pref(X + ) (resp. Suf(X + )) stands for all nonempty prefixes (resp. suffixes)<br />

<strong>of</strong> X + .<br />

Now we can state:<br />

Conway’s Problem. Is the centralizer <strong>of</strong> a regular X regular as well?<br />

Although the answer is believed to be affirmative, it is known only in the<br />

very special cases, namely when X is a prefix set, binary or ternary, see [15], [3]<br />

or [7], respectively. This together with the fact that we do not know whether<br />

the centralizer <strong>of</strong> a finite set is even recursive, can be viewed as an evidence <strong>of</strong><br />

amazingly intriguing nature <strong>of</strong> the problem <strong>of</strong> commutation <strong>of</strong> languages.<br />

Example 1. (from [3]) Consider X = {a, ab, ba, bb}. Then, as can be readily seen,<br />

the centralizer C(X)equalstoX + \{b} =(X∪{bab, bbb}) + . Hence, the centralizer<br />

is finitely generated but doesn’t equal either to X + or {a, b} + .<br />

Finally, we note that in the above the centralizers were defined with respect<br />

to the semigroup A + . Similar theory can be developed over the free monoid A ∗ .<br />

3 Fixed Point Approach<br />

As discussed extensively in [14] and [8], there has been a number <strong>of</strong> different<br />

approaches to solve the Conway’s Problem. Here we introduce one more, namely<br />

so-called fixed point approach. It is mathematically quite elegant, although at<br />

the moment it does not yield into breakthrough results. However, it can be seen<br />

as another evidence <strong>of</strong> the challenging nature <strong>of</strong> the problem.<br />

Let X ⊆ A + be an arbitrary language. We define recursively<br />

X0 =Pref(X + ) ∩ Suf(X + ), and<br />

Xi+1 = Xi \ [X −1 (XXi∆XiX) ∪ (XXi∆XiX)X −1 ], for i ≥ 0, (1)<br />

where ∆ denotes the symmetric difference <strong>of</strong> languages. Finally we set<br />

Z0 = �<br />

Xi. (2)<br />

We shall prove<br />

i≥0<br />

Theorem 1. Z0 is the centralizer <strong>of</strong> X, i.e., Z0 = C(X).<br />

Pro<strong>of</strong>. The result follows directly from the following three facts:<br />

(i) Xi+1 ⊆ Xi for all i ≥ 0,<br />

(ii) C(X) ⊆ Xi for all i ≥ 0, and<br />

(iii) Z0X = XZ0.

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