without context – The language $ {ww | w in {0,1 } ^ {*} } $ n is not a CFL

We proved that the language $ L = { Omega Omega | omega in {0,1 } ^ {*} } $ It's not a CFL, and we did it using the pumping lemma. And the proof is clear to me. But I thought about the following CFG:

$ G = ( {S, S_ {1} }, {0,1 }, R, S) $ where R has the following rules:

$ S rightarrow S_ {1} S_ {1} | epsilon $

$ S_ {1} rightarrow 0S_ {1} | 1S_ {1} | epsilon $

It seems that the language of this CFG should be the language $ L $ that I've defined above since each substitution adds the same letter on both sides. But it can not be so since we can use the pumping lemma on the word $ 0 ^ {l} 1 ^ {l} 0 ^ {l} 1 ^ {l} $ (or $ l $ is the pumping length). So, either I do not do the substitution incorrectly, or the CFG language contains $ L $ and has more words than I currently do not see …

Can any one help me and point out where is my mistake?

Conversion of CFG to CFL [on hold]

Is it possible to convert
(1) Each CFL CFL
(2) Each CFL at CFL


If we take a string where the number of a and the number of b equals the number of c
then we get a CFG that does not have CFL.
So, can CFG at CFL be possible or not?

CFL and LED mixed

I shoot a video on a green screen. I need to turn on the screen as well as the subject. I have a bunch of 6000K CFL and I want to get some LEDs. If both types have the same temperature, 6000 K, can I mix them in a video or will that be a problem?

Prove that each CFL is decidable in a space O (n)

This question was raised while a group of students from my school were studying for our qualifying exams. The question on an old exam was:

Prove that each language without context $ A $ is in $ mathrm {SPACE} (n) $. You can assume that $ A $ is given in CNF (normal form of Chomsky)

I've seen the CYK algorithm but its spatial complexity is $ O (n ^ 2) $.