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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.
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A context free grammar G = (N, Σ, P, S) is in binary form if for all productions A we have |α| ≤ 2. In addition we say that G is in Chomsky Normaml Form (CNF) if it is in bi
Construct a Mealy machine that can output EVEN or ODD According to the total no. of 1''s encountered is even or odd.
What are codds rule
We now add an additional degree of non-determinism and allow transitions that can be taken independent of the input-ε-transitions. Here whenever the automaton is in state 1
A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note
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We represented SLk automata as Myhill graphs, directed graphs in which the nodes were labeled with (k-1)-factors of alphabet symbols (along with a node labeled ‘?' and one labeled
s->0A0|1B1|BB A->C B->S|A C->S|null find useless symbol?
We will assume that the string has been augmented by marking the beginning and the end with the symbols ‘?' and ‘?' respectively and that these symbols do not occur in the input al
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