0

I'm currently trying to answer a question were I have to confirm if a language is regular or not.

If the language is not regular I have to give an informal answer to why the language is not regular but i'm not required to do any proofs such as pumping lemma.

I've been asked to identify some aspect of the language which which cannot be modeled by either an automaton, right-linear grammar or left-linear grammar.

Given the following example: $\{ba^n\cdot bc^n \mid n \ge 1\}$

how can I informally say why this isn't a regular language.

Thanks in advance.

BLAZE
  • 1
user5525615
  • 15

1 Answers1

0

Here’s a semi-formal explanation; you could probably loosen it up a bit and still get the idea across.

A right-linear grammar would generate words of the language from left to right, so at some point in the derivation of $ba^nbc^n$ it would have generated $ba^nbX$ for some non-terminal symbol $X$. Since the only word of the language that begins with $ba^nb$ is $ba^nbc^n$, $c^n$ must be the only terminal string that can be generated from $X$. There are only finitely many non-terminal symbols, so there are only finitely many positive integers $n_1,n_2,\ldots,n_m$ for which there are non-terminal symbols that generate $c^{n_1},c^{n_2},\ldots,c^{n_m}$. If $n>\max\{n_1,\ldots,n_m\}$, there is no non-terminal symbol in the grammar that generates $c^n$, so the grammar cannot generate $ba^nbc^n$ and therefore does not generate the given language.

By the way, you can find a fairly careful proof of this using the pumping lemma here.

Community
  • 1
Brian M. Scott
  • 631,399