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

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?

I could make one definition with more As and one with more Bs, but then the number of characters is still correlated — Florian Ludewig, Oct 27 '19 at 12:27
I don't understand your objection. Inequality is also a form of correlation. — rici, Oct 27 '19 at 12:37
Ok, maybe I misunderstood your hint. Could you further elaborate on how to actual implement it? — Florian Ludewig, Oct 27 '19 at 16:22
But you yourself suggested the solution above ;-) You just need to finish it. — rici, Oct 27 '19 at 17:19
Possible duplicate of Context Free Grammar for language $L={a^ib^j \mid i,j \ge 0; i \ne 2j}$ — Yuval Filmus, Oct 27 '19 at 18:37
See https://cs.stackexchange.com/questions/10387/cfg-using-the-set-l-left-a-b-right-for-the-expression-ambn-whe?noredirect=1&lq=1 — Yuval Filmus, Oct 27 '19 at 18:37
@YuvalFilmus thanks. Its not exactly what I need, but through it, I figured it out :) — Florian Ludewig, Oct 27 '19 at 20:11

score 3 · Accepted Answer · answered Oct 28 '19 at 08:05

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: $$ \begin{align} &S \to A \mid B \\ &A \to aAb \mid aA \mid a \\ &B \to aBb \mid Bb \mid b \end{align} $$

All of your solutions produce the sentences $a^i$ and $b^i$ for $i\ge 1$, although the OP requests $a^ib^j, i,j\ge 1, i\ne j$. Of course, the fixes are minor. For the first one, you can change $T\to aTb\mid\epsilon$ to $T\to aTb\mid ab$. — rici, Oct 28 '19 at 20:27

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

1 Answers1