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What is the grammar for language $L = \{ a^nb^m : n\neq m-1\}$?

I only know I have to write grammar for both $ n<m-1 $ and $ n>m-1 $, so this is what I wrote:

  • For $n<m-1$: \begin{align} &S\to Abb \\ &A\to aAb \mid \lambda \end{align}

  • For $n>m-1$: \begin{align} &S\to aaA \\ &A\to aAb \mid \lambda \end{align}

Yet I can not mix them and get a correct grammar for $L$.

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  • $\begingroup$ What makes you think that there's a context-free grammar for it? $\endgroup$
    – Raphael
    Commented Jan 19, 2018 at 13:43
  • $\begingroup$ @Raphael As the question seems to distinguish $<$ and $>$ it seems we have to read $!$ as negation, rather than as factorial. $\endgroup$
    – Hendrik Jan
    Commented Jan 19, 2018 at 14:18
  • $\begingroup$ PS. I took the liberty of editing both $!=$ and landa. $\endgroup$
    – Hendrik Jan
    Commented Jan 19, 2018 at 14:21
  • $\begingroup$ @HendrikJan Ahhh, good point. Thanks! $\endgroup$
    – Raphael
    Commented Jan 20, 2018 at 0:38

2 Answers 2

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If you split the grammar into two parts for disjoint parts of the language you better take different names for the variables, since they have a different meaning in the two parts.

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In your first grammar, change S to X and leave A unchanged. In your second grammar, change S to Y and A to B. Then add S -> X | Y.

And fix your grammars to handle n < m-1 instead of n=m-2 and n>m-1 instead of n = m+2.

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