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  Spontaneous Transitions, Finite Automata and Regular Languages

Spontaneous transitions: convenient, however not essential

Lemma (ε-transitions):

Any NFA N with the ε-transitions is transformed to an equivalent NFA M devoid of ε-transitions.

The fundamental idea is simple, as described below. The ε-transitions from r to s implies that anything which N can do starting in s, like an a-transition from s to t, N can already do beginning in r. After adding suitable new transitions, like the a-transition from r to t, the ε-transition from r to s can be erased.

949_lemma transitions.jpg

Definition: the ε-closure E(s) of a state s is a set of states which can be reached from s following ε-transitions.

In illustration above: E(r) = {r, s}, E(s) = {s}, E(t) = {t}. The importance of this concept is due to the fact that if N reaches s, it may as well reach any state in E(s) devoid of reading any input.

The given illustration describes the general construction which transforms a NFA N with ε-transitions (at left) to an equivalent NFA M with no ε-transitions (at right).

Illustration: L = {0i 1j 2k | i,  j, k  ≥ 0} = 0*1*2*.  This language is usual of the structure of communication protocols. The message comprises of a prefix, a body and a suffix, in this order. When any part might be of arbitrary length, involving zero length, the language of legal messages consists of the structure of L.

699_Illustration of NFA.jpg

Substitute each and every state s of N by the state E(s) of M: E(s0) = {s0, s1, s2}, E(s1) = {s1, s2}, E(s2) = {s2}. This reflects the fact that if N got to s0, it may spontaneously have proceeded to s1 or to s2. E(s0) is the beginning state of M. Any state E(s) of M which includes an accepting state of N should be made an accepting state of M - in our illustration; all the states of M are accepting! Lastly, we adjust the transitions: As E(s0) = {s0, s1, s2}, M’s  transition function f assigns to the f( E(s0), 2) the union of all N’s transitions outgoing from s0, s1, s2: f( s0, 0), f( s1, 1), f( s2, 2).

In spite of the fact that all M’s states are accepting, M only accepts the strings in the L. The first symbol ‘2’ of string ‘21’, for illustration,  leads M from E(s0) to E(s2), however E(s2) can’t process the second symbol ‘1’, therefore ‘21’ is not accepted.

Illustration: NFA N transformed to an ε-free NFA M. E(s1) = {s1, s3}.  L = (a ∪ ba* (a ∪ b) a)*

2407_NFAN transformation.jpg

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