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I'm trying to allow an empty transition in a PDA for the following language:

  • Alphabet: $Σ = \{a, b, c\}$
  • Language: $L = \{ a^ib^j \mid i \neq j \} \cdot \{ c \}^\ast$

Examples of words in $L$:

  • $\varepsilon$
  • $aabccc$
  • $abbccc$

Not in $L$:

  • $abcc$
  • $aabbc$

Here is what I came up with:

PDA

The diagram above uses JFLAP - where the symbol $Z$ reflects the empty stack. The symbol $λ$ is the empty symbol $ϵ$.

It accepts everything as it should, but I don't know how to let epsilon get through. q7 to q8 is when there is more b than a. So there should be a way to allow q7 to q9 where a is more than b but also epsilon can get through. Thoughts? I would like to simply set epsilon through but than aabbc can get through easily enough.

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  • $\begingroup$ With letting "epsilon get through" do you mean accepting the empty word? $\endgroup$ – dkaeae Feb 1 at 15:31
  • $\begingroup$ @dkaeae Sorry yeah, meant accepting the empty word $\endgroup$ – Andrew Raleigh Feb 1 at 15:41
  • $\begingroup$ @Apass.Jack You're right, fixed to clarify $\endgroup$ – Andrew Raleigh Feb 1 at 15:43
  • $\begingroup$ Is the $x,y;z$ notation on the state transitions supposed to mean the PDA reads input symbol $x$ and stack symbol $y$ followed by writing $z$ to the stack? Then why is there a symbol $Z$ in the transition from $q_7$ to $q_8$ being read from the stack (if no such symbol was pushed previously)? $\endgroup$ – dkaeae Feb 1 at 16:18
  • $\begingroup$ Is $Z$ used to symbolize the end of the stack? Do you use $\lambda$ as $\epsilon$? $\endgroup$ – phan801 Feb 1 at 16:28
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Some key points:

  • The only case where you want to accept $\epsilon$ is when you haven't had any other input and your stack is empty. Therefore you need an accepting state right at the beginning.
  • You need to take into consideration inputs such as: $bac, aac,...$ that will never result in an accepting state, or $abbb$ that results in an accepting state without containing any $c$'s
  • If you get a $b$ and your stack is empty you have more $b$'s than $a$'s so you need an accepting state
  • If you get a $c$ and and your stack is not empty you have more $a$'s than $b$'s so you need an accepting state
  • You reject everything else.

This works

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  • $\begingroup$ Didn't think it would be this complicated, thanks! I do question how $λ, Z : Z$ works? Wouldn't this mean we can put anything in (i.e. a) and it would then accept it? If that's the case would an epsilon transition between $q7$ and $q9$ work as well? $\endgroup$ – Andrew Raleigh Feb 1 at 19:52
  • $\begingroup$ I'm sorry, you're right! You don't need $\lambda,Z : Z$ at all. You just enter at $q_0$ which is an accepting state. I'll edit $\endgroup$ – phan801 Feb 1 at 19:58
  • $\begingroup$ Another question, from $q4$ to $q5$, does having the $λ, λ:λ$ transition mean anything can go there? Would that mean you can sneak in an $a$ in there and get it accepted? $\endgroup$ – Andrew Raleigh Feb 1 at 20:23
  • $\begingroup$ You're right, once ageain. See the edit and let me know if something else doesn't seem right $\endgroup$ – phan801 Feb 1 at 20:42

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