Performing ϵ-closure on NFA

33 Views Asked by At

Say you have an NFA where the only transitions from the start state are $\epsilon$, when performing an $\epsilon$-closure then you have a state that is completely isolated and I'm not sure if my understanding is correct, because the result doesn't make sense to me. As an example:

a b c ε
-> A BC *
B B B *
C C C *

Where the $\epsilon$-closure table is:

$\epsilon$-closure
-> A ABC *
B B *
C C *

My understanding then is that after performing an $\epsilon$-closure, we get this NFA

a b c
-> A *
B B B *
C C C *