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Is $L / R$ context free?

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I was reviewing the post If $L$ is context-free and $R$ is regular, then $L / R$ is context-free?

I completely understand why $L/R$ is context free.

I just tried a different approach, which is not that rigorous according to the answers given. It is just based on the basic understanding.

How I tried :

Let $L = a^N b^N a^M b^M$ and $R = \left (a + b \right )^*$.

Then $wx$ belongs to $L$. Suppose I assume $wx = aaabbbab$.

So, now $x = ab$ which belongs to $R$.

So $L/R$ contains $aaabbb$.

Like this, I will get all strings in $L/R = \left \{ a^N b^N | N \geq 1 \right \}$.

This language is nothing but context free.

This approach is not as rigorous as what I saw in the answers.

I just want to confirm: is this approach right?

Asked By : Garrick
Answered By : Yuval Filmus

You can't prove a general theorem by proving one special case. So even if your proof of the special case were correct (and it isn't), all it does is prove that the theorem holds in that specific case.

Your proof is erroneous since $L/R$ is larger than what you indicate. In fact, whenever $\epsilon \in R$, we always have $L/R \supseteq L$. In your case $$ L/R = \{ a^{n_1} b^{n_2} : n_2 \leq n_1 \} \cup \{ a^n b^n a^{m_1} b^{m_2} : m_2 \leq m_1 \}. $$

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