[Solved]: Non-regular Languages?

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How to prove that a language is not regular?

Why $L_a$ and $L_b$ are not reguluar?

$L_a = \{ e^i f^{n-i} g^j h^{n-j} : n \in N, 1 \leq i, j \leq n \}$.

$L_b= \{nm^{i_1} nm^{i_2}...bn^{i_z}: z \in N, (i_1,...,i_n) \in N^z, 1 \leq j \leq z, i_j ≠ j \}$.

Asked By : corium

Answered By : David Lewis

Hints...

$L_a$ can be transformed into an well-known non-regular language by a very simple homomorphism that basically "loses" information.

For $L_b$ you might try transforming it into its "opposite" (that is, equality in place of inequality) by operations known to preserve regular languages. The "opposite" language might then be amenable to a pumping argument. Also, try "testing" your argument on a much simpler version of $L_b$ which, although it is regular, will let you see the patterns to use for the full $L_b$.

PS -- these are both techniques that might be added to the reference post that Dave Clarke points to. I will do so when this question has settled down.

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Thinking a bit more about $L_b$, the idea I had in mind to derive the "opposite" language may not lead directly to an easy solution. So here's a hint in a different direction. There are a lot of $m^{i_j}$ in each string of that language. If you could "isolate" those strings of $m's$ by an appropriate regularity-preserving operation, you might get a simpler language that is amenable to a proof, perhaps by pumping or perhaps by the "opposite" idea. Also, think about what each $i_j \neq j$ means in terms of other symbols in the string.

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Question Source : http://cs.stackexchange.com/questions/1753