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.


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What do you think about these developments?

Ekran Görüntüsü (1467).png
Ekran Görüntüsü (1469).png

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

Please help me with this question.
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?enter image description here

Pumping Lemma Doubt (LONGERB)- Theory of Computation

Please help me with this question.
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?

finite automata – Using pumping lemma to show that there is always a smaller word for a regular language

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.

L is a regular language in this case.

I understand its saying that for any string in a regular language there can be a shorter version of it to an extent. But I’m not sure how to begin using the pumping lemma to show this. I have used the pumping lemma to show something is not regular, but I haven’t done a proof like this before…