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I need to understand the epsilon transition for NFA, for example I have an NFA with four states q0, q1, q2 and q3 the latter is an accepted state.

In particular the transition function is:

{q0, a}= {q1}
{q1, eps}= {q2}
{q2, b}= {q3}

I would implementing this behavior in java.

I have the string 'ab' and I have to see if it is accepted. I begin reading the character 'a' and arrival in the state q1, the next character is 'b' but the state q1 don't have transition for 'b'.

I have to run the epsilon transition as I read the character 'a'? I do not know how to do this in my application, can you help me?

Thanks

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  • $\begingroup$ questions asking for mere implementation in some specific language are off topic here. $\endgroup$ – sashas Apr 29 '16 at 14:38
  • $\begingroup$ what topic should I add? but logically the epsilon transition I have to read it after reading the letter 'a' or before reading the letter 'b'? thank you very much $\endgroup$ – stella Apr 29 '16 at 14:51
  • $\begingroup$ You do the epsilon transition without having read any input $\endgroup$ – NaCl Apr 29 '16 at 18:22
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You seem to be misconstruing what an $\epsilon$-transition is. If we have, say, a transition rule $\delta(q_1,\epsilon)=\{q_2\}$ it means that the FA can switch from state $q_1$ to $q_2$ without reading any input. In your example, on input $ab$ the FA reads the $a$ and so passes from $q_0$ to $q_1$. It can then (in one reality) stay in $q_1$ or (in another), pass to state $q_2$. Then, on seeing the $b$, there is indeed no move from $q_1$ (in one reality) but there is a move (in another reality) to $q_3$, so since the NFA reaches a final state in one reality, the input $ab$ is accepted by the NFA.

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  • $\begingroup$ Thank's for your answer. My question is the transition δ(q1,ϵ)={q2} It must be made before the reading 'b' or after reading 'a'? Thank for all $\endgroup$ – stella May 2 '16 at 11:14
  • $\begingroup$ @stella. In your example, you can regard the two as identical. $\endgroup$ – Rick Decker May 2 '16 at 12:35

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