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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.
1. Does above all''s properties can be used to prove a language regular? 2..which of the properties can be used to prove a language regular and which of these not? 3..Identify one
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Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with
We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while
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First model: Computer has a ?xed number of bits of storage. You will model this by limiting your program to a single ?xed-precision unsigned integer variable, e.g., a single one-by
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