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I am studying for an exam and came up to this task in the picture

usually the transformation from a NFA to a DFA is easy to me. I create a transition table from the NFA and for every "new" combined state, i create a new column and so on (looking at my try explains it better)

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Somehow I am skeptical with the single state of q3. Since no state can reach single q3, does this even make sense?

ArdianH101
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  • I follow everything and get virtually the same answer as you, but without q3 in the DFA. Why did you put q3 in there? There is no path through the NFA that can lead exclusively to q3, is there? It's not wrong to put all 2^|Q| states in the DFA, some are just unreachable or otherwise useless, as {q3} is in your DFA. Why not put {q2} in there in that case, though? And {}? Etc. – Patrick87 Aug 26 '19 at 12:18

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Your DFA is correct. It contains an unnecessary and unreachable state q3, but this is perfectly fine; not all states need to be reachable. Unreachable states would be removed during minimization, but the subset construction does not attempt to guarantee that the resulting DFA will be minimal.

The way I run the subset construction, I only include states corresponding to subsets that can be reached; but it would not be incorrect to simply include all possible subsets. It's just a stylistic choice, and even my method does not yield minimal DFAs, only ones without unreachable states.

Patrick87
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