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If my alphabet is {a,b} and my nfa has the following transitions:

State    |      a        b        epsilon        
--------------------------------------------
q0              q1      null        q1
q1              q2       q1         none
q2              q2       q1         none

Is this table wrong? should delta(q0, b) = q1 because q0 can move on epsilon to state q1?

kjh
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  • I think you are talking about epsilon-transitions right? ;) – olydis Sep 14 '13 at 19:50
  • Yes, I'm just not sure how to write that into my question as a symbol, I'll edit it to make it more obvious – kjh Sep 14 '13 at 19:51
  • if `delta(q0, b) = q1` is wrong depends on transitions in state q1! – olydis Sep 14 '13 at 19:52
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    If from q1 (or any state you reach from q1 via epsilon-transition) you reach q1 **when b is read**, **then** `delta(q0, b) = q1` holds for your DFS ;) – olydis Sep 14 '13 at 19:54
  • Ah so basically because delta(q0, epsilon) = q1, then that means delta(q0, b) = delta(q0, epsilon) is that correct? – kjh Sep 14 '13 at 19:54
  • what should "1" mean as second argument to `delta`??? – olydis Sep 14 '13 at 19:55
  • I mistakenly mis-typed that comment, edited just now – kjh Sep 14 '13 at 19:56
  • okay, but no, that is wrong ;) – olydis Sep 14 '13 at 19:57
  • I see no reason why the transition table you posted should be wrong... as said, **if** it is wrong as you suggested, it would require `delta(q1, b) = q1` ... which I do not know since you did not paste the transition-table of **all** states – olydis Sep 14 '13 at 19:58

1 Answers1

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As olydis stated in the comments section:

"If from q1 (or any state you reach from q1 via epsilon-transition) you reach q1 when b is read, then delta(q0, b) = q1"

kjh
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