LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

128 Karel Culik II, Juhani Karhumäki, and Petri Salmela
final states states transitions
X0 6 8 15
X1 5 9 17
X2 6 13 24
X3 9 19 35
X4 12 25 46
X5 15 31 57
X6 18 37 68
X7 21 43 79
X8 24 49 90
X9 27 55 101
X10 30 61 112
X11 33 67 123
X12 36 73 134
X13 39 79 145
X14 42 85 156
X15 45 91 167
X16 48 97 178
X17 51 103 189
X18 54 109 200
X19 57 115 211
final states states transitions
X20 60 121 222
X21 63 127 233
X22 66 133 244
X23 69 139 255
X24 72 145 266
X25 75 151 277
X26 78 157 288
X27 81 163 299
X28 84 169 310
X29 87 175 321
X30 90 181 332
X31 93 187 343
X32 96 193 354
X33 99 199 365
X34 102 205 376
X35 105 211 387
X36 108 217 398
X37 111 223 409
X38 114 229 420
X39 117 235 431
Table 1. The numbers of states, final states and transitions of automata corresponding
to the iteration steps for the language X.
What we can conclude is the following much weaker result, first noticed in [8]:
Theorem 2. If X is recursive, then C(X) is in co-RE, that is its complement
is recursively enumerable.
Proof. As we noticed above, all approximations Xi are recursive, and moreover
effectively findable. Now the result follows from the identity
C(X) = �
Xi,
where bar is used for the complementation. Indeed, a method to algorithmically
list all the elements of C(X) is as follows: Enumerate all words w1,w2,w3,...
and test for all i and j whether wi ∈ Xj. Whenever a positive answer is obtained
output wi.
i≥0
For regular languages X we have the following result.
Theorem 3. Let X ∈ A + be regular. If C(X) is regular, even noneffectively,
then C(X) is recursive.
Proof. We assume that X is regular, and effectively given, while C(X) isregular
but potentially nonconstructively. We have to show how to decide the membership
problem for C(X). This is obtained as a combination of two semialgorithms.