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I have been asked to find a grammar that will generate the language $\{a^{n^2}:n \ge0\}$ in an exercise. So far I tried to replicate the previously written characters with my grammar rules but it didn't work. Any idea on how to setup such grammar? Any help will be appreciated.

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Found the answer here. Basically the grammar looks like this: $$ S→LAYR \\ ZA→aAZ \\ Za→aZ \\ ZR→AAYR \\ aY→Ya \\ AY→YA \\ LY→LZ \\ YR→X \\ aX→Xa \\ AX→Xa \\ LX→ε \\ $$

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    $\begingroup$ And what's the idea? Why would that be correct? $\endgroup$ – Raphael Mar 15 '17 at 21:53
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Set aside $n$ of the $n^2$ characters, and use that $(n+1)^2 = n^2 + 2n +1$.

So at any moment we have (say) $n^2-n$ symbols $A$ and $n$ symbols $B$. We also need endmarkers. $L,R$ count as A$.

$S\to LBBR$

In one linear phase we visit all symbols, and each symbol $B$ will generate two extra $A$'s. Finally we add a $B$.

$L \to LX$; $XA\to AX$; $XB\to AABX$; $XR\to BR$

Nondeterministically end, transforming all symbols into $a$.

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  • $\begingroup$ This is a bit sketchy... $\endgroup$ – Yuval Filmus Mar 15 '17 at 19:12
  • $\begingroup$ @YuvalFilmus OK. added details. $\endgroup$ – Hendrik Jan Mar 15 '17 at 21:21

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