state-machine
is this Epsilon-NFA correct?
I am exercising and wasn't sure if I got this correct. I had to draw 0* U 1*.
Update: This was correct
Solution
yh you are right , This NFA accepts either 0 or 1
so you can say NFA for 0 U 1
as epsilon concatenate with 1 results in 1
or epsilon concatenate with 0 results in 0
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