I was working through an NFA to DFA conversion tutorial: where I'm converting the following NFA to a DFA:
NFA to a DFA:
I used lazy subset construction which I'm sure the tutorial did as well because we had the same transition table of:

delta      |  0        |  1       |
{a}        |{a,b,c,d,e}|{d,e}
{d,e}      |{e}        |{∅}       
{b,d,e}    |{c,e}      |{e}       
{e}        |{∅}        |{∅}       
{c,e}      |{∅}        |{b}       
{b}        |{c}        |{e}
{c}        |{∅}        |{b}

However, there state diagram for the resulting DFA is this:
Resulting DFA

I'm confused as to why the set {b,d,e} isn't being labeled as an accepting state.

Also, I don't know if I'm allowed to ask this but if there is a place with more NFA to DFA converting practice I'd love to know, right now I'm just using any old study guide I can find on google.


1 Answer 1


You're right, {b, d, e} should be a final state. Otherwise the DFA would reject 0001 while the "equivalent" NFA accepts it.

  • $\begingroup$ That makes me feel better about my understanding, thanks $\endgroup$ Commented Feb 7, 2017 at 20:43

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