Show that a language cannot be generated by linear grammar

Question

I have a language $ L= \{ w \in \{a,b\}^* ; |w|_b=2i, i \ge 0 \}$ that is a language with even number of b's.

I found a grammar for it with these rules:

$S \rightarrow aS \ | \ bL \ | \ \lambda $

$L \rightarrow aL \ | \ bS \ $

How could I show that this language cannot be generated by linear grammar?

According to Wikipedia, a linear grammar is a context-free grammar that has at most one nonterminal in the right hand side of its productions.

Sorry, my lecturer had bad defintion in slides, I edited it, But still, this language is not regular, and I think its not linear too. — Martin, Apr 22 '16 at 13:10

score 2 · Answer 1 · edited Apr 13 '17 at 12:48

2

Your grammar is right-regular and can thus be used to prove that $L \in \mathrm{REG}$. Since every regular grammar is linear, your claim is impossible to prove.

edited Apr 13 '17 at 12:48

Community

1

answered Apr 22 '16 at 12:25

Raphael

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My intuition says this language is not regular, so I'm trying to prove its not linear either ( regular $\ne$ linear ) – Martin Apr 22 '16 at 13:04
@martinerk0 Your intuition is wrong, and contradicts your statement "I found a [regular] grammar for [$L$]". – Raphael Apr 22 '16 at 13:29
Why would that grammar be regular? – Martin Apr 22 '16 at 13:31
It's right-regular; just check the definition. It's trivial to transform it into a DFA. – Raphael Apr 22 '16 at 13:32

score 1 · Answer 2 · answered Apr 22 '16 at 12:32

1

A grammar whose only rules are of the form $X \to uYv$ and $X \to w$, where $u,v,w$ are terminals, can only generate words of odd length. Your language also contains words of even length. If you also allow rules of the form $X \to \epsilon$, then you can construct a grammar for your language (exercise).

answered Apr 22 '16 at 12:32

Yuval Filmus

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I apologize, now I have correct definition of linear grammar. – Martin Apr 22 '16 at 13:10

Show that a language cannot be generated by linear grammar

2 Answers2