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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.
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
4 bit digital comparator png
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
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Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
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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
One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
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