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How to prove that a language is not regular?

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

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

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

corium
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    Have you tried the techniques listed here: http://cs.stackexchange.com/questions/1031/how-to-prove-that-a-language-is-not-regular? – Dave Clarke May 09 '12 at 14:05
  • The fact that there is no descriptive title shorter than the question shows that this is a bad question. What have you tried? Why do the standard approaches not work? Closed as duplicate for now; if you add specific information, you can vote/flag to reopen. – Raphael May 09 '12 at 16:02
  • @Raphael -- Can you point to the earlier question(s) that are exact duplicates? Thanks. – David Lewis May 10 '12 at 02:09
  • @DavidLewis: This is automatically done above the question whenever a question is closed as duplicate. Do not take the phrasing "exact duplicat" too literally, though; it is just that the linked question answers the question completely as it stands. – Raphael May 10 '12 at 07:23
  • @Raphael -- sorry, did not realize that link was there. I'm not sure that link does discuss the techniques for these particular problems. As I said, I will add an answer there covering the additional ideas, since that seems to be a reference post. – David Lewis May 10 '12 at 14:07
  • @DavidLewis: The other question contains a reference to closure properties; what more do you use? – Raphael May 10 '12 at 14:10
  • @Raphael -- more detail and examples on closure properties, which is a diverse subject: homomorphism, complement, $\cap$ regular set, transducer, arbitrary quotient, reversal and combinations of all these. (There are also Kleene closure, catenation and union, though they tend not to help non-regularity proofs.) For example, I think $L_b$ in this example needs complement, homomorphism and perhaps transducer and quotient, at least the method I have in mind. – David Lewis May 10 '12 at 17:36
  • @DavidLewis Ok, I see how such an answer could enrich the reference question (which contains only a link to closure properties now). Please go ahead! – Raphael May 12 '12 at 08:54

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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.


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.

David Lewis
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  • Can you give me hints for the homomophism? - for $L_a$ i tried $h(a) = x, h(b) = x, h(c) = y, h(d)= y$ than i have ${x^{ij}y^{(n-i)(n-j)} :n \in N, 1 \leq j,j \leq n}$ this language is non-regular! - is the way right? – corium May 09 '12 at 17:44
  • If by $a, b, c, d$ you mean, in order, $e, f, g, h$ then you are absolutely on the right track. But it looks like you calculated the resulting language wrong. Try straightening out your symbols carefully -- things should get simpler, not more complex. – David Lewis May 09 '12 at 20:23
  • See addendum about $L_b$. – David Lewis May 10 '12 at 02:22