Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. If $G$ is not a regular grammar, then $L(G)$ is infinte.

  2. If $L^*$ is context free then $L$ is definitely context free.

  3. If $G$ is a context free grammar that is language is $L$ (meaning $L(G) = L$),

    then there exists a context free grammar $G^r$ such that $L(G^r) = L^r$

    • $L^r =\{w^r \mid w \in L\}$
share|improve this question
    
Please make the title more descriptive. Thanks. –  Babak S. Dec 21 '12 at 14:13
    
@BabakSorouh I went ahead and changed the title a bit as part of the task of cleaning up all of the spelling mistakes. user14988, please check to see that I didn't alter the meaning of your question with my grammar tweaks. –  rschwieb Dec 21 '12 at 14:20
    
@rschwieb: Yes, and I’ve made the change. –  Brian M. Scott Dec 21 '12 at 14:24
    
What does "definitely context free" mean? Note that for any language $L$, $(\{0,1\} \cup L)^* = \{0,1\}^*$ is context free. –  Carl Mummert Dec 21 '12 at 14:35

1 Answer 1

This isn’t a full answer, but it’s too long for a comment.

(1) is actually false as stated. The grammar $G$ with initial symbol $S$, terminal symbols $0$ and $1$, non-terminal symbols $S,A,B,C$, and the productions below is clearly not regular, but $L(G)=\{01\}$.

$$\begin{align*} &S\to ABC\\ &A\to 0\\ &B\to 1\\ &C\to C\epsilon C\mid\epsilon \end{align*}$$

What is true is that if $L$ is a finite language, then there is a regular grammar $G$ such that $L(G)=L$; this is probably what you were really meant to prove.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.