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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.
We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
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
#can you solve a problem of palindrome using turing machine with explanation and diagrams?
Construct a Mealy machine that can output EVEN or ODD According to the total no. of 1''s encountered is even or odd.
Theorem The class of recognizable languages is closed under Boolean operations. The construction of the proof of Lemma 3 gives us a DFA that keeps track of whether or not a give
conversion from nfa to dfa 0 | 1 ___________________ p |{q,s}|{q} *q|{r} |{q,r} r |(s) |{p} *s|null |{p}
When we study computability we are studying problems in an abstract sense. For example, addition is the problem of, having been given two numbers, returning a third number that is
let G=(V,T,S,P) where V={a,b,A,B,S}, T={a,b},S the start symbol and P={S->Aba, A->BB, B->ab,AB->b} 1.show the derivation sentence for the string ababba 2. find a sentential form
Design a turing machine to compute x + y (x,y > 0) with x an y in unary, seperated by a # (descrition and genereal idea is needed ... no need for all TM moves)
How useful is production function in production planning?
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