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The language accepted by a NFA A = (Q,Σ, δ, q0, F) is
NFAs correspond to a kind of parallelism in the automata. We can think of the same basic model of automaton: an input tape, a single read head and an internal state, but when the transition function allows more than one next state for a given state and input we keep an independent internal state for each of the alternatives. In a sense we have a constantly growing and shrinking set of automata all processing the same input synchronously. For example, a computation of the NFA given above on ‘abaab' could be interpreted as:
This string is accepted, since there is at least one computation from 0 to 0 or 2 on ‘abaab'. Similarly, each of ‘ε', ‘ab', ‘aba' and ‘abaa' are accepted, but ‘a' alone is not. Note that if the input continues with ‘b' as shown there will be no states left; the automaton will crash. Clearly, it can accept no string starting with ‘abaabb' since the computations from 0 or ‘abaabb' end either in h0, bi or in h2, bi and, consequentially, so will all computations from 0 on any string extending it. The fact that in this model there is not necessarily a (non-crashing) computation from q0 for each string complicates the proof of the language accepted by the automaton-we can no longer assume that if there is no (non-crashing) computation from q0 to a ?nal state on w then there must be a (non-crashing) computation from q0 to a non-?nal state on w. As we shall see, however, we will never need to do such proofs for NFAs directly.
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
A finite, nonempty ordered set will be called an alphabet if its elements are symbols, or characters. A finite sequence of symbols from a given alphabet will be called a string ove
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
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c program to convert dfa to re
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DEGENERATE OF THE INITIAL SOLUTION
spam messages h= 98%, m= 90%, l= 80% non spam h=12%, m = 8%, l= 5% The organization estimates that 75% of all messages it receives are spam messages. If the cost of not blocking a
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
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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