by languages are the same. – For M. 1., M. 2., L(M. 1.) = L(M. 2.) • DFAs and NFAs: – For every NFA there is an equivalent DFA. (we’ll prove this) and vice-versa. Problem Statement. Let X = (Qx, ∑, δx, q0, Fx) be an NDFA which accepts the language L(X). We have to design an equivalent DFA Y = (Qy, ∑, δy, q0, Fy) such. To show this we must prove every DFA can Consider the NFA that accepts binary strings ending with The key idea for building an equivalent DFA is to.
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By using our site, you acknowledge that dfs have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. It is true that this construction may result in a DFA with unreachable states. The general construction begins simply by including all possible states, then adding the appropriate transitions, so typically the resulting DFA won’t be the smallest DFA that accepts the same language in terms of the number of states.
An alternative approach is to only add states as you generate the transitions rather than adding all states at the start. This will give you only reachable states, but even then, this DFA may not be the smallest possible.
If you indeed follow the construction in the way you describe, then there might be states which are unreachable from the starting state. That’s allowed in a DFA, though you can go ahead and remove them without affecting the operation of the automaton.
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Home Questions Tags Users Unanswered. Paresh 2, 1 14 For a proof it doesn’t matter is there are extra states, or too many states; what matters it that it is easy to write and understand the proof.
For practical use such considerations are or central impportance, and complicate things quite a bit. Summarizing, as always theory is easier than practice.
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Explain with example equivalence between NFA and DFA
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