Pumping Lemma (For Regular Languages)

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my professor watches the same lecture 5 min prior to the class

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thank you for repeating some parts. it really helps me capture what you're saying.

[@cvit47]

These videos will pump you up!

[@AdarshNinganurthebest]

5. Method to prove that a language L is not regular  At first, we have to assume that L is regular.  So, the pumping lemma should hold for L.  Use the pumping lemma to obtain a contradiction o Select w such that |w| ≥ n o Select y such that |y| ≥ 1 o Select x such that |xy| ≤ n o Assign the remaining string to z. o Select k such that the resulting string is not in L.  Hence L is not regular.

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I just love the way u teach sir... It helped me a lot...thank you

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Awesome lecture and great music at the end.

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Great lectures,your explanation is in a huge way,we are understanding very clearly and also provide gate lectures sir for theory of computation

[@ccppurcell]

In the statement of the lemma, it should be that S is a member of A. "If A is reg, there exists integer P such that all S in A with length at least P can be divided into xyz with these properties..."

[@JamateSirajiaFaisalabad]

String S is any word of language A

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better than my professor several times...

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I really loved my professor for the class I took in which this is one of the topics, but he seriously stumbled over this explanation even though he explained everything else really well

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This is absolutely brilliant and you explain it in such an easy way!!!!

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really i love u Sir because u are the best lecturer i have ever seen thanks thanks. please complete for us the subject

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Im watching this video 6 years after taking theory of computation and finally able to understand this.

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Man, what an explanation👍🏻👌🏼👌🏼👌🏼👌🏼 this is the 15th video I'm watching in a row and never liked this many videos in less time😁😁👌🏼👌🏼👍🏻👍🏻👍🏻

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i can not afford to buy paid your course. i am in 3rd sem now btech. pleaseeeeee dont remove these lectures. its is extremely helpfull.

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Thanks for the clear explanation.

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for the sake of us , you are doing a great job sir, hat's of you sir, i would like to know your name sir...

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A man has no name XD

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+Hrishikesh parab hahaha

[@jithapnair2641]

Thank you for your explanation 😊

[@horizonpenblade1288]

So how do you know what P is for a given language? I mean, I know I'm misunderstanding something here but it feels like this theorem as it is could be used to prove a regular language irregular by using the wrong pumping length.

[@nekdo9590]

You never know the exact value of P. The only thing you know is that such number EXISTS.

[@oooBASTIooo]

Essentially P is the smallest number of states a machine for that language can have + 1. So you usually do not know it, unless you know the minimal automaton. You usually have to say, given P, .... so u dont get to chose the P, as you have to argue against EVERY possible P.

[@rajatprajapati6386]

best lecture in youtube sir

[@Amy-tw3zh]

This is wrong. You don't get to choose pumping length p. You also don't get to choose how to "split" the string. You only get to choose the string s and i.

[@phoenixgamming9346]

pls add the video with example..

[@aishikbhattacharjee3180]

Awesome explanation. But can you please what is pumping length?

[@Omar-kw5ui]

its the length of how much you expanded your regex. say you have a* this can be aaa or aaaaa or aaaaaa. the pumping length is how many times you expanded a. Thats the best explanation I can give from the tutorial.

[@harishr5620]

@@Omar-kw5ui It's not the pumping length..Pumping length is defined as the minimum number of states required b the automaton to accept the finite language.

[@JamateSirajiaFaisalabad]

@@Omar-kw5ui right

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Those here for exam tomorrow ! 🔥

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👋🏻🙂

[@muhammadshehrozaziz4058]

what is the pumping lenght? is it given ?

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By far the best video

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Most toughest subject in the world

[@ziliestarrive]

Why do you have to show that all three conditions cannot be satisfied to have a contradiction? Isn't proving that any one of the three condition cannot be satisfied enough?

[@JamateSirajiaFaisalabad]

Yes any one condition is enough to prove a language non regular.

[@zerobit778]

Can we say if we could not use pumping lemma to prove "Not regular", then it is regular? Suppose regular language is a set, then the complement set is non-regular language. If a language is non-regular we could use pumping lemma to prove. If a language is regular we could not use pumping lemma to prove(2:44). Thus If we could not use pumping lemma, then it would be regular. It's same like somewhere wrong, but I could not find it. plz help~

[@Wock__]

No, the pumping lemma cannot be used to prove a language is regular. I believe there are non-regular languages that satisfy the pumping lemma, therefore this is just one method we can use to prove certain languages are non-regular.

[@ehteshamulhaq604]

aa(ab)*bb is Regular Language??? Can we prove it as non regular using Pumping Lemma?

[@somtara3515]

that's quite helpful! thx!

[@elenafeder3923]

wait so what is it called when a language is neither regular nor non-regular?

[@Lelio1111]

Thanks a lot!

[@krishnasaivadapalli619]

ambn for m,n>=1 is regular or not . Has anybody tried Pumping Lemma for this? Plz explain

[@lamaspacos]

We find a string S and make all their possible divisions.

[@aliyyah82]

Very helpful video :)

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Could you provide your slide material pdf or ppt version?

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appriciable sir... can u please upload an example of this also...

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thanks very nuch this chanel good for this course

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Shout out to our professor who doesn't even teach us this lesson and let us watch NESO Academy's video instead, I'm so tired of you, I'm so annoyed!!! anyway thanks to NESO

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LMAO🤣 Yoohh I'm here for the same reasons!!

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i think my teacher is also following this TOC playlist. she is providing your content as it is of you in class. Is this a scam?

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Thanks ! You are Great !!!

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Very Useful

[@rameshch8698]

what is exactly p...how do you know that value.....

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That was helpful. Thanks

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Thanku ♥️

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This guy is awesome!!

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sir can you please tell the book u take reference from?

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Thank you..Sir..

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Just watched one minute of this before realizing it's not about English grammar.

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Thank you sir😊

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Great Work . Some links in ur website are not working. Please Fix them.

[@origin3100]

If the length of v is equal to 0 isnt it against the second point of the th?

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Best explanation ever 👌👌

[@oromis995]

An example would have been nice, but good explanation

[@mohammadvasimbaig8680]

00:04 Pumping lemma is used to prove that a language is not regular. 01:07 The pumping lemma can be used to prove that a language is not regular. 02:09 The language can be divided into three parts: X, Y, and Z 03:16 To determine if a language is regular, three conditions from the pumping lemma must be satisfied. 04:16 To prove that a language is not regular 05:12 We need to find a string in language A such that its length is greater than or equal to the pumping length and then divide it into three parts X, Y, and Z. 06:18 The string s cannot be formed as a regular language. 07:15 Using the pumping lemma, we prove that a language is not regular. Crafted by Merlin AI.

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Who are here before exam watching many tutorials but understand nothing

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tomorrow is my paper and you are my saviour ty sir

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Thanks a lot

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Plz upload data compression lecture sir

[@lM4nu]

Gracias.

[@dhanushsivajaya1356]

Thankyou sir

[@radiaIes]

Keine sorge, wir packen Hollas schon.

[@bdjeosjfjdskskkdjdnfbdj]

Intuition for why this lemma is true?

[@hashrajput5528]

is P (pumoing length) denotes prime?

[@JamateSirajiaFaisalabad]

No it is not prime It is named as pumping length For example: L = {a^n b^n} where n>0 Here pumping length is n

[@Prachita516]

was your teacher Saurabh Shanu?

[@bilalchandiabaloch8464]

Sir L={ww|w €(a+b)* is a regular or Non regular language according to pumping lemma plz make a video on this language. Thanks in anticipation

[@FleurDeCersier]

L is not a regular language. Let's choose for instance s=(b^p)a(b^p)a=xyz. If i=0 (or anything but 1), then s=xy^0z=(b^p-1)a(b^p)a which proves that L is not regular according to pumping lemma because p-1 != p Tried to explain it the best I could but I study CS in my native language which is not English. Hope you got the gist though

[@M-ABDULLAH-AZIZ]

a^n b^ n where n+m is even , how is this a regular language? one can prove it is not regular by pumping lemma, but I read online it is regular. Please help me understand this!

[@chinmayhattewar4456]

we can form regular expressions for this string i.e. it is recognizable by FSM, Hence it is regular.

[@hrishikeshbhaskar3161]

Sir pls make a video on pigeon hole principle

[@govandibotelo2841]

I have a Theory of computation exam in a few hours 😅

[@amolbambode2959]

am i right in saying that case2 and case3 do not follow property 3 of pumping lemma? |xy|

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in the next video

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Explanation and representation is great! #thumbsUp

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thanku sir

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ur great sir

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Video very good

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Good

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Thanks for teaching.

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I am really sad for on this incredible day, I did not understand what was going on in this video. Therefore consider this a formal complaint!

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30 min before exam😂

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Hail Neso!!!

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?