Pumping Lemma (For Regular Languages) | Example 1

TOC: Pumping Lemma (For Regular Languages) | Example 1

This lecture shows an example of how to prove that a given language is Not Regular using Pumping Lemma.

Music:
Axol x Alex Skrindo - You [NCS Release]

[bestof]

I am ready to PUMP some bleach into my stomach and bid this cruel world goodbye. Thanks theory of Computation

[nasirmehmood]

I think we are not suppose to proof second and third condition as |y| >0 & |xy|<=p. We have to use these as our base conditions and on the bases of these conditions we have to proof that xy^n is not a regular language.

[chaturthi]

Please continue doing this great work sir ...it means a lot..

[rayansailani]

Just to clarify things. the explanation is correct in the sense that he checks for the |xy|<=P, But we must also check the other conditions as well. Yes we could have stopped there also, since this was the first example sir wanted to check for all the cases and make us understand. The pumping length of 7 could have been any arbitrary value. Something not too small and not too large to not cause confusions and for our better understanding. So YEAH!

[thomaswillems]

Thank you so much, this is so much clearer than how my professor teaches the pumping lemma, great job!

[mustafaalan]

Giving a value to P is not a correct solution for most of the University Exams and not an exact solution. Instead please give K+L+M to P. [ x=a^K, y=a^L, z=a^M+b^P ] Then when you increase the i which is on top of y, you will see that on the left side there will be L more a's according to rule |y|>0 . This is how you prove that it is not a regular language. (Case 2 and 3 are cancelled by the Rules 2 and 3. That's why there is no point to check those cases)

[todarfclips]

Very nice example and very nicely explained. Thanks !

[unzilechris]

This is great, thank you. I now understand how the pumping lemma works.

[alexrubio]

You a hard topic incredibly simple to understand, thank you

[SadmanAmin]

Why we are going for case 2 and 3? They already broke the rule of |xy|<=P!

[asspsu]

YOU ARE SO GOOD!!!! I love the way how you explain it very clearly and slow down the proof!! Works on a no native English speaker like me!

[Theraverguy]

Great videos thank you pls continue the great work

[cyeganeh]

dude you fucking saved me for my final man. my teacher and tas could not teach this for their life, and your 2 videos on pumping lemmas explained what they could not explain. They had 2 and a half hours and you did it in 20 minutes. Thank you.

[hugonguyensg]

The part where you actually define p=7 is potentially misleading because p is just a symbolic threshold and will be useful in cases such as proofing inequalities by utilizing the 2nd and 3rd conditions in pumping lemma. So we don't necessarily need to know what is p. This example just need to use the 1st condition of the lemma to proof.

[inowhy]

You are very good at explaining things! You must know that :) Thanks for the help!

[zihengtang]

the idea is that p is an abstract length, given that length of (xy) must be equal or less of p, y can only consist of letter(a)s, since the rule is a^pb^p. howeve given y is all letter(a)s, xy^iz for all i>=0 is not in the language, thus that it is not regular for any p value

[yuriileso]

In the 2nd and 3rd case you have broken 2th PL rule.
I would recommend to use some parameters, not numerical values
F.E. case 2:
Let say, that PL length is just p;
So, we chose the next string: a^p.b^p.
x = a^t.b^s
y = b^n
z = b^i

The 2nd rule of PL tells, that |xy|<=p, so t+s+n should be less or equal to p;
Our string tells that t := s+n+i (the numbers of "a" is equal to numbers of b).
t+s+n <= p && t = s+n+i >= p. As the result we can use t value on 1th equation.
t + s + n = (s + n + i) + s + n := 2s + 2n + i.
we know t (:= s+n+i) is more or eq. to p; so 2s+2n+i is sure more than p.
Here is the contradiction we a looking for. We have broken 2th PL rule

[catcen]

THANK YOU SO MUCH!! YOU ARE THE BEST!!

[moeyali]

The pumping lemma was pumping me because my professor didn't teach it well. However you taught this so well

[ajinkyawandale]

In case 2 |xy| = 13 and in case 3 |xy| = 9 which violates the condition of pumping lemma which states that |xy| should be less than of equal to pumping length ( here 7 )