I'm trying to find a grammar of the highest type possible for this language:
L={0^2n 1^(n-1)|n>=1}
I only managed to do this:
S->00X
X->00X1|λ
Which is not type 3. I can't seem to figure out how to get it to type 3 (if that's even possible).
You can't do it, because L is not a regular language.
Assume that L is regular. Let w = 0^(2p)1^(p-1) for some integer p>=1, so that |w| > p. Further, consider the strings x, y, and z such that w = xyz with |xy| <= p, which means both x and y are sequences of 0s (since p < 2p). By the pumping lemma, any string of the form xy^nz is also in L, but that means we can increase the number of 0s without increasing the number of 1s found in z. Thus, our assumption that L is regular must be false.