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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:
In general non-determinism, by introducing a degree of parallelism, may increase the accepting power of a model of computation. But if we subject NFAs to the same sort of analysis
Automaton (NFA) (with ε-transitions) is a 5-tuple: (Q,Σ, δ, q 0 , F i where Q, Σ, q 0 and F are as in a DFA and T ⊆ Q × Q × (Σ ∪ {ε}). We must also modify the de?nitions of th
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For every regular language there is a constant n depending only on L such that, for all strings x ∈ L if |x| ≥ n then there are strings u, v and w such that 1. x = uvw, 2. |u
Prepare the consolidated financial statements for the year ended 30 June 2011. On 1 July 2006, Mark Ltd acquired all the share capitall of john Ltd for $700,000. At the date , J
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
We will specify a computation of one of these automata by specifying the pair of the symbols that are in the window and the remainder of the string to the right of the window at ea
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
Computer has a single LIFO stack containing ?xed precision unsigned integers (so each integer is subject to over?ow problems) but which has unbounded depth (so the stack itself nev
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