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It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ v) directly computes another (p, v) via a path that includes some number of ε-transitions (before the σ-transition, after it or both), we can get the same effect by extending the transition relation to include a σ-transition directly from q to v. So, in the example we could add ‘a' edges from 0 to 1 (accounting for the path 0 3) and from 1 to 3 (accounting for the path 1 3) and ‘b' edges from 1 to 3 (accounting for the path 1 3), from 3 to 2 (accounting for the path 3 2), and from 1 to 2 (accounting for the path 1 2), Note that in each of these cases this corresponds to extending δ(q, σ) to include all states in ˆ δ(q, σ). The remaining effect of the ε-transition from 0 to 2 is the fact that the automaton accepts ‘ε'. This can be obtained, of course, by simply adding 0 to F. Formalizing this we get a lemma.
Myhill graphs also generalize to the SLk case. The k-factors, however, cannot simply denote edges. Rather the string σ 1 σ 2 ....... σ k-1 σ k asserts, in essence, that if we hav
what exactly is this and how is it implemented and how to prove its correctness, completeness...
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Exercise Show, using Suffix Substitution Closure, that L 3 . L 3 ∈ SL 2 . Explain how it can be the case that L 3 . L 3 ∈ SL 2 , while L 3 . L 3 ⊆ L + 3 and L + 3 ∈ SL
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Distinguish between Mealy and Moore Machine? Construct a Mealy machine that can output EVEN or ODD According to the total no. of 1's encountered is even or odd.
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