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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 regular expressions rather than just symbols in Σ∪{ε}. We will explain the algorithm using the example of Figure 1.
We begin by adding a new start state s and ?nal state f to the automaton and by extending it to include an edge between every state in Q∪{s} to every state in Q ∪ {f}, including self edges on states in Q. We then consolidate all the edges from a state i to a state j into a single edge, labeled with a regular expression that denotes the set of strings of length 1 or less leading directly from state i to state j in the original automaton. If there was no path directly from i to j in the original automaton the label is ∅. If there were multiple edges (or edges labeled with multiple symbols) the label is the ‘+' of the symbols on those edges (as in the edge from 2 to 1 in the example). There will be an edge from s labeled ε to the original start state and one labeled ∅ to every other state other than f. Similarly, there will be an edge labeled ε from each state in F in the original automaton to state f and one labeled ∅ from those in Q-F to f. The expression graph for the example automaton is given in the right hand side of the ?gure.
The idea, now, is to systematically eliminate the nodes of the transition graph, one at a time, by adding new edges that are equivalent to the paths through that state and then deleting the state and all its incident edges. In general, suppose we are working on eliminating node k. For each pair of states i and j (where i is neither k nor f and j is neither k nor s) there will be a path from i to j through k that looks like:
Question 2 (10 pt): In this question we look at an extension to DFAs. A composable-reset DFA (CR-DFA) is a five-tuple, (Q,S,d,q0,F) where: – Q is the set of states, – S is the alph
Give the Myhill graph of your automaton. (You may use a single node to represent the entire set of symbols of the English alphabet, another to represent the entire set of decima
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Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no
A.(A+C)=A
All that distinguishes the de?nition of the class of Regular languages from that of the class of Star-Free languages is that the former is closed under Kleene closure while the lat
program in C++ of Arden''s Theorem
We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also
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What are the issues in computer design?
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