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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:
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De?nition Instantaneous Description of an FSA: An instantaneous description (ID) of a FSA A = (Q,Σ, T, q 0 , F) is a pair (q,w) ∈ Q×Σ* , where q the current state and w is the p
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Give DFA''s accepting the following languages over the alphabet {0,1}: i. The set of all strings beginning with a 1 that, when interpreted as a binary integer, is a multiple of 5.
Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
Another striking aspect of LTk transition graphs is that they are generally extremely ine?cient. All we really care about is whether a path through the graph leads to an accepting
What is the purpose of GDTR?
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
short application for MISD
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