This channel is criminally underrated! I cant emphasize how much this comic style awesome, adorable and clean is in its explanations. Thank you very much and keep up the good work!
when u take 3 cases, in those last 2 cases, you are taking y such that |xy| is already greater than p. You basically need to chose a y such that |y| is greater than 1 and less than p and |xy| must be less than p.
Yet another top quality video! From animation to explanation and the funny art style to the soothing voice its all just prefect! This channel is so underrated :'(
You definitely should open your own university. and then you should just sit there as a tutor and play these vids... you will be a tutor of the year for sure. :) thanks for the great explanations. helped me a lot in my exams.
Thank you very much. I really liked the video. Honestly a kinda underaprecciated channel now that i look at the subs and views. You deserve more. Keep it up.
Lydia! You're the best! Thank you so much for your work! I would happy to see videos of yours about some problems like determining if the language is finite/ empty etc. Thanks again!
Sorry, I was so distracted by the beautiful sound of your voice - couldn't focus on what you were saying at first :-) But seriously, a great video, very intuitive and with nice metaphors like the "all-knowing Pumping Lemma" on its throne. Thanks for this!
At 4:30, I assume you wanted to say : Assume the language is regular. Then given P, we find a string that can't, in fact, be pumped and still be in the language. This proves the language is non-regular, which contradicts our initial assumption.
In regular language we can make state diagram, so we know, number of states=p but in non-regular language we cannot make state diagram for language. then how we are going to find p for non-regular language?
Hi I had a question regarding the 3 cases for the pumping lemma. If we could prove one of the cases was true for the y value chosen, and it met all the three conditions for pumping lemma, then that would show the language is regular then, correct?
The language you’re talking about is different though. She was proving that the language consisting of an uninterrupted sequence of n 0s followed by an uninterrupted sequences of n 1s is irregular. Strings in such format correspond to stuff like 000111, 01, 00000001111111 and so on, while 01010101, for example, doesn’t belong to such language.
Again, this is higher quality than a majority of the classes I have taken so far. Thank you so much.
That makes two of us.
@@aloafofbagels6381 And three
4 in mutual agreeance
5 in one accord
6 weary seekers of truth
5-minute concise explanations are so much better than lectures that are over an hour long. Much easier to remember :)
Literally summed up like 2 hours of my lecture in 5 minutes thank you so much
This is exactly what I needed before my teacher had us retake the test.
This channel is criminally underrated! I cant emphasize how much this comic style awesome, adorable and clean is in its explanations. Thank you very much and keep up the good work!
Your videos explain content better than my whole semester of class. Thank you.
when u take 3 cases, in those last 2 cases, you are taking y such that |xy| is already greater than p. You basically need to chose a y such that |y| is greater than 1 and less than p and |xy| must be less than p.
can you explain this more please
This saved me. Thank you so much. Sharing with the rest of my class right now.
This guy doesn't understand how to get a cheeky curve
@@RagingAcid lmao that's fuckin evil
This is great! Wish there was more videos where you did example question for tricky problems as well!
Thank you!! your voice is beautiful and easy to understand :)
I wish you still did these sort of video!
Thank you so much for this clear explanation and accompanying animation! Very clear and concise.
Yet another top quality video! From animation to explanation and the funny art style to the soothing voice its all just prefect! This channel is so underrated :'(
notice how non-regular language is so chill bout not being a regular language bro has a positive attitude
love your videos tho just subscribed
wonderful video ...waiting for context free languages
DUDE YOU ARE A GODDESS, THANK YOU
This video is excellent, it simply explains this confusing topic
You're litteraly saving me on Computing Theory
You definitely should open your own university. and then you should just sit there as a tutor and play these vids... you will be a tutor of the year for sure. :)
thanks for the great explanations. helped me a lot in my exams.
Thank you very much. I really liked the video. Honestly a kinda underaprecciated channel now that i look at the subs and views. You deserve more. Keep it up.
Thank you! Your videos are exceptional :)
I have been on TH-cam since as log as I can remember,I can say this is TOP-NOTCH content 🔥.
this video may have just saved me for my midterm omg
Wow! This was a great proof indeed!
Lydia! You're the best! Thank you so much for your work! I would happy to see videos of yours about some problems like determining if the language is finite/ empty etc.
Thanks again!
well, i dont need a pumping lemma to proof you are not regular, amazing! for basic understanding ofcourse
Fantastic video, thank you so much! Really made it much easier to understand proving non-regularity and my understanding of the pumping lemma too :)
Thank you. You are smarter than Neso Academy
I am really in love with your videos, please make more! (I'm spreading the words :D )
Thanks so much!
Truly helped more than any other sources I've encountered.
Great explanation
I love your video! Please keep it up!
very clear, love the pictures and the flow. thank you.
These animations are soooo good. Please make a whole series on Automata! TYSM
Beautiful playlist. Loved it
Thank you so much; this was amazing! Just gave you your 1,000th like
Wonderful video!
This is worth a lot! Thank you!
Thank you so much. This was very helpful and I was also confused in this part but now this is very clear.
best explanation ive seen, thanks!
Please increase the video audio quality next time. It's hard to hear through the speaker. Otherwise, the video is very helpful thank you!
you have saved my life! thanks
These videos are so helpful, thank you!
Can you please make a video about regular expressions and more on proofs please these have been very helpful
amazing vid! The textbook or professor does not do this concept any justice lol
Sorry, I was so distracted by the beautiful sound of your voice - couldn't focus on what you were saying at first :-) But seriously, a great video, very intuitive and with nice metaphors like the "all-knowing Pumping Lemma" on its throne. Thanks for this!
This was super helpful thank you so much!
I wonder if we choose the string 0^p1^p and choose y to be 1^p. Doesn’t it not satisfy the condition of |xy| =< p? 😃
Super good explanation!!!!
Brilliant! Thank you!
At 4:30, I assume you wanted to say : Assume the language is regular. Then given P, we find a string that can't, in fact, be pumped and still be in the language. This proves the language is non-regular, which contradicts our initial assumption.
In regular language we can make state diagram, so we know,
number of states=p
but in non-regular language we cannot make state diagram for language.
then how we are going to find p for non-regular language?
To my humble knowledge, we don't necessarily choose P as the number of states as it is dependant on the language and not the machine.
thank u very much
better than automata course that i take in the college.
I hate that this video is 2 years old. Please make more videos!
how to determine what length p to use?
Great video
1:07 What about finite regular languages? They don't satisfy the pumping lemma but are still considered regular.
How do you define p? I know it is a pumping length but how you define it when you don't know which length is a pumping string?
Thanks! :>
Thankssssss
very well done :)
10/10 video
3:13 how are cases 2 and 3 possible if |xy|
Hi I had a question regarding the 3 cases for the pumping lemma. If we could prove one of the cases was true for the y value chosen, and it met all the three conditions for pumping lemma, then that would show the language is regular then, correct?
amazing. what am I paying my professor for
Loved this!
Couldnt you have chosen a pair of 01, and thus keep the sequence?
010[10]1->010[10101010]1
Yes but I think you need to find at least one sequence that is not part of the language
The language you’re talking about is different though. She was proving that the language consisting of an uninterrupted sequence of n 0s followed by an uninterrupted sequences of n 1s is irregular.
Strings in such format correspond to stuff like 000111, 01, 00000001111111 and so on, while 01010101, for example, doesn’t belong to such language.
😍
But the volume was too low...
pushing p
This rules
Ty so much