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Let A = {(B,w) | B is an NFA that accepts w}. To prove that N is a decidable language,

M = "on input (B,w) where B is NFA and w is a string, simulate B on w. If B accepts w, accept. Otherwise, reject"

Is this a false proof? The proof from the textbook says that I have to convert NFA into equivalent DFA but I'm not sure why this would be necessary. Is it not possible to simulate NFA directly without conversion?

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  • $\begingroup$ Do you know how to simulate a NFA on w? How would you do that? How would you determine whether it accepts or not? $\endgroup$
    – D.W. ♦
    Dec 4, 2017 at 3:35
  • $\begingroup$ I haven't thought about it but I guess if deterministic TM can implement non-deterministic TM, why not NFA? $\endgroup$
    – Ted
    Dec 4, 2017 at 19:30

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The textbook probably suggests you convert into a DFA because it is easier to simulate a DFA than to simulate a NFA and determine whether the DFA accepts the word w. This is trickier on a NFA (but still possible, if you know how).

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  • $\begingroup$ Thanks. Also, does "simulating" something mean that the TM is going to run it (like provoking a function call inside a function), or TM is going to mimic the behaviour of that machine? $\endgroup$
    – Ted
    Dec 4, 2017 at 3:41

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