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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.
The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
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
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
1. An integer is said to be a “continuous factored” if it can be expresses as a product of two or more continuous integers greater than 1. Example of continuous factored integers
We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also
1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization
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 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
The Equivalence Problem is the question of whether two languages are equal (in the sense of being the same set of strings). An instance is a pair of ?nite speci?cations of regular
automata of atm machine
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