## Prove with pumping theorem that the language { \$a^n b^n b^m a^m | n ≠ m \$ } is not context free

I’m having a trouble proving it to be context free, with an example if I take w = $$a^k b^k b^{k+1} a^{k+1}$$

then it would be problematic if the partition of $$vxy$$ with $$|v| = |y|$$ was in the $$b^{k+1} a^{k+1}$$ part as I can’t get it to be equal with $$k$$ or make the amount of $$a$$ and $$b$$ different

It is also allowed to be proven with Ogden’s Lemma

## Pumping lemma for an involved non context free language

Hi I’m trying to show $$C={wzzw^R|w,zin{0,1}^+}$$ is not a context-free language.( I have this believe because $$C={ww|win{0,1}^+}$$ is not a context free language.) I’m really struggling to come up with a string that captures the essence of irregularity of this language: I tried strings like $$s=1^p0^p1^p0^p1^p1^p$$ but there are too many cases to deal with and most of the examples I saw only use 1 or 2 cases, so I believe the direction I’m going is wrong. Can you provide a hint on which string to pick as the ‘pumping string’? Thank you.

## Use pumping lemma (non regular language) to solve

{0^m 1^n 0^m | m,n >=0}

vv^R :v: {a,b}*

## regular languages – Why does the Pumping Lemma Constraint |xy| ≤ p mean that y can’t be 1 in the string 0p1p

I am trying to get my head around the Pumping Lemma to prove a language is non-regular.

I am reading the Sipser text book and he gives the following example.

Let B be the language $${0^n 1^n | n ge 0}$$

Let $$s = 0^p 1^p$$

I understand that the idea is you can split this string into xyz such that y can be pumped. It is the constraint of $$|xy| le p$$ that is confusing me.

Sipser notes that due to this constraint y could not equal 01 nor could it equal 1. Why would y equaling either of those values violate the given constraint.

I am generally quite confused by the Pumping Lemma so any general advice or good resources you can recommend, I would appreciate.

Thanks!

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## regular languages – How to proove L 2^n with Pumping lemma

Let $$Sigma$$ is the latin alphabet ({a,b,c…,x,y,z} – 26 letters).

Given language
$$L = { alpha in Sigma^{*} |$$

if $$alpha$$ cointaints $$a$$ then $$N_{a}(alpha) = 4$$

if $$alpha$$ cointaints $$b$$ then $$N_{b}(alpha) = 8$$

$$quad vdots$$

if $$alpha$$ cointaints $$b$$ then $$N_{z}(alpha) = 2^{27}$$

Prove that L is reagular.

Note: I did research and I have found that we can use Pumping lemma

## regular languages – Pumping lemma: why x in ∣xy∣ ≤ p?

Looking at the pumping lemma, I’ve noticed that in the string $$xy^pz$$, there seems to be no rule explicitly stated for $$x$$ and $$z$$. If I understand correctly, $$x$$ and $$y$$ are basically anything on the 2 sides of the string $$y^p$$ that we’re pumping and thus can be anything in $$L$$.

Rule 2 & 3 of the pumping lemma are:

• $$|y| geq 1$$
• $$|xy| leq p$$

Since $$|x| = 0$$ and $$|z| = 0$$ seem to be allowed, as they only need to be of non-negative length, we shouldn’t need $$x$$ in rule 2 and it can be rewritten as $$1 leq |y| leq p$$.

Are $$x$$ and $$y$$ not just a substitute for whatever are on the 2 sides of $$y^p$$ which we’re pumping? Why is $$x$$ in rule 2 if it doesn’t seem to make a difference? If $$x$$ is necessary, why is there no $$|yz| leq p$$?

## regular languages – Pumping Lemma Doubt (LONGERB) – Theory of Computation

The answer that has been given in the book is 3 (Options 1,2, and 4). But my doubt is, that the 4th option doesn’t belong to the given language as there are equal number of a and b in this option. So, according to me the correct answer should be 2 (Options 1 and 2 only).
Please give a proper reason for each of the four options, why it can or can not be used to prove that the given language is not regular?

## Pumping Lemma Doubt (LONGERB)- Theory of Computation

I’m having trouble putting together a mathematical proof that uses the Pumping Lemma to show that $$exists$$n $$geqslant$$ 1 such that that for all strings w $$in$$ L such that |w| $$geqslant$$ n, there is another string z $$in$$ L such that |z| < n.