Context-free grammar for $\{a^x b^y : x \neq y\}$

Question

I am trying to create a context free grammar in Extended Backus–Naur form, which starts with a non-empty sequence of a's and is followed by a non-empty sequence of b's. With the special condition that the number of b's has to be unequal to the number of a's.

Thus, the grammar should generate words like:

• aaaabbb
• aaabb
• abbb

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So basically I could do something like this:

$\ G=(N,T,P,S)$

$\ N = \{S\}$

$\ T = \{a,b\}$

$\ P = \{S=aa(S|\epsilon)b\}$

But then the words would always have $\ 2n$ a's and n b's:

• aab
• aaaabb
• aaaaaabbb

So how is it possible to make the number of a's uncorrelated of the number of b's, without being equal?

Answer 1

Let $L = \{a^n b^n : n \in \mathbb N\}$. Your language can be written as $a^+L \cup Lb^+$, and this leads to the following grammar: $$ \begin{align} &S \to AT \mid TB \\ &T \to aTb \mid \epsilon \\ &A \to aA \mid a \\ &B \to bB \mid b \end{align} $$ We can save a nonterminal by factoring $L$ differently: $$ L = \{a^na^mb^n : n \geq 0, m \geq 1\} \cup \{a^nb^mb^n : n \geq 0, m \geq 1\}. $$ This leads to the following grammar: $$ \begin{align} &S \to aSb \mid A \mid B \\ &A \to aA \mid a \\ &B \to bB \mid b \end{align} $$ There are many other possible variants, for example: $$ \begin{align} &S \to A \mid B \\ &A \to aAb \mid aA \mid a \\ &B \to aBb \mid Bb \mid b \end{align} $$