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I am wondering if this is even possible, since $\{a^n b^n c^n \mid n \geq 0\} \not\in \mathrm{CFL}$. Therefore a PDA that can distinguish a word $w\in\{a^n b^n c^n \mid n \geq 0\}$ from the rest of $\{a^*b^*c^*\}$ might as well accept it, which sounds contradictory to me.

I guess I need to take advantage of the non-deterministic nature of PDAs but I'm out of ideas. If you could offer some advice I would very much appreciate it.

Juho
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hauptbenutzer
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  • Interesting point about it seeming contradictory. Indeed, context-free languages are not closed under taking the complement... so there are lots of examples of non-context-free languages that could be "accepted" in the sense you allude to. I'm not a theorist and, as such, can't really reconcile this, but perhaps someone else can chime in on why this isn't something to worry about? – Patrick87 Dec 05 '12 at 18:45
  • Note that this generalizes: the complement of ${a^n b^n c^n d^n e^n}$ is a CFG. – sdcvvc Dec 05 '12 at 19:41
  • Similar question. – Raphael Dec 06 '12 at 06:04
  • Is not the title of this question wrong? E.g. the complement of ${a^n b^n c^n | n \geq 0}$ is not ${a^b^c^* - {a^n b^n c^n | n \geq 0}}$. – O. Altun May 30 '20 at 13:13

1 Answers1

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No, this is context-free. To accept $a^nb^nc^n$, you need to make sure that three numbers are equal. To accept $a^*b^*c^* \setminus a^nb^nc^n$, you just need to make sure that you're in one of the following three cases:

  1. The number of $a$s is different from the number of $b$s; or
  2. The number of $a$s is different from the number of $c$s; or
  3. The number of $b$s is different from the number of $c$s.

Write a PDA for each of these cases, then combine them by jumping nondeterministically to each one from the start state.

Patrick87
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  • I'd written down these cases alright, but I was missing the idea to connect them. Thank you! – hauptbenutzer Dec 05 '12 at 18:42
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    Actually you need only any two cases. – sdcvvc Dec 05 '12 at 19:39
  • @sdcvvc Good point. :) – Patrick87 Dec 05 '12 at 20:24
  • For different number of characters, consider this as inspiration: $S → xSy | X | Y ; X → x | xX ; Y → y | yY$. It should be simple to glue either $a^+$ onto the left of this or $c^+$ onto the right and turn this into a PDA. For the tricky case (which you don't need) $S → aSc | A | C ; A →aB | aA ; C → Bc | Cc ; B → ε | bB$. – Jonas Kölker Jul 31 '16 at 12:02