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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:
Trees and Graphs Overview: The problems for this assignment should be written up in a Mircosoft Word document. A scanned hand written file for the diagrams is also fine. Be
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what is regular expression?
For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable. "Positiveness Problem". Note that
The class of Strictly Local Languages (in general) is closed under • intersection but is not closed under • union • complement • concatenation • Kleene- and positive
Another way of interpreting a strictly local automaton is as a generator: a mechanism for building strings which is restricted to building all and only the automaton as an inexh
Computer has a single FIFO queue of ?xed precision unsigned integers with the length of the queue unbounded. You can use access methods similar to those in the third model. In this
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
Prove that Language is non regular TRailing count={aa ba aaaa abaa baaa bbaa aaaaaa aabaaa abaaaa..... 1) Pumping Lemma 2)Myhill nerode
Applying the pumping lemma is not fundamentally di?erent than applying (general) su?x substitution closure or the non-counting property. The pumping lemma is a little more complica
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