thank you so much for every video of toc i reallly appriciate your work man.we are much more better than my teacher.i am watching your video for my toc paper.thank you so much you just saved me from failing.
I learn more from watching UA-cam videos than my four years in college. Pretty much, you can replay the videos if you ever have a hard time understanding.
thank you so much for every video of toc i reallly appriciate your work man.we are much more better than my teacher.i am watching your video for my toc paper.thank you
You said if the 3 conditions are true then it is a context free language. That’s not necessarily the case. The claim is that IF a language is context free then the pumping lemma must hold for that language. It is possible for a non context free language to have those properties.
@@kelvinvu1247 No this is not true. If any, not all, of the three conditions are violated, it is not a context free language. What OP is saying is correct and is not the same as what you are saying. Essentially, these 3 conditions are necessary but not sufficient for proving a language is context free, they can only show you definitively that a language is NOT context free.
if its pumpbable at least in one of the division possibilities then the contradiction proof fails. To understand this better you have to walk slowly through the thought process of the proof: We ASSUME that the language is pumpable. Therefore, it MUST HAVE a pumping length, call it P, so that ALL words longer than P MUST BE PUMPABLE. From here, we set to prove the contradiction: since our claim has led to saying ALL words longer than P are pumpable, it's enough to find One word that can't be pumped to prove the contradiction. If you picked a word and it turns out to be pumpable, you should try to find a different word that won't be; if you cant find one, it might be the case that the language is actually context free (it's not guaranteed, pumping lemma doesnt prove anything)
CFL closed under concatnation is valid but L1={a^n, n>0} L2={b^nc^n, n>0} L1.L2={a^nb^nc^n} --> whic is not CFL Then How it is closed under CFL? Please help me........................
Well, instead of using the same index 'n' both times maybe using the index 'm' in L2 will make this more clear to you why the concatenation is still a CFL.
This 8 minutes lecture is far better than my 1 hour professor lecture. Thanks a lot.
@Hendrix Quinton Yup, been watching on Flixzone for since november myself :)
@Hendrix Quinton Yea, I've been watching on flixzone for months myself :D
Concentration matter brother
so true, most professors seem to hate giving lectures
Absolutely right 👍❤
Thank you so much for making these videos.
Tomorrow we have an exam and we will watch yours video until morning, you have really saved our asses.
Hope your ass is still safe after 4 years 😂
@@JustwaitNwatch-w oh i think you are also here to save yours
@@Quran_ver of course dude 🤣
@@Quran_ver my ass was not saved but after rechecking the marks i was passed in this stupid subject
This is great. You are amazing. Never understood this with my professor's slides.
That's because you never paid attention
@@nirdeshpathak4332 bruh!! he paid enough attention to this subject that's why he came here to understand the topic ! you dumb or wot?😒😒
@@nirdeshpathak4332 That's such a cap
thank you so much for every video of toc i reallly appriciate your work man.we are much more better than my teacher.i am watching your video for my toc paper.thank you so much you just saved me from failing.
I learn more from watching UA-cam videos than my four years in college. Pretty much, you can replay the videos if you ever have a hard time understanding.
I'm a non cs student, and never studied CS cncepts in college, but youtube videos pretty much covers it all
Neso academy is not good channel it is one of the best channel for TOC
I love this guy, he has helped me out so many times
Thank you for this; it's very helpful!
method of understand is very well. full personalizims. thanks sir
thank you so much!, i hope that will help me to pass my exam tomorrow!
You will bro...BEST OF LUCK...
thank you so much for every video of toc i reallly appriciate your work man.we are much more better than my teacher.i am watching your video for my toc paper.thank you
You said if the 3 conditions are true then it is a context free language. That’s not necessarily the case. The claim is that IF a language is context free then the pumping lemma must hold for that language. It is possible for a non context free language to have those properties.
this is correct
right, basically you need to find a contradiction for ALL 3 conditions based off a string S, to prove it is not context free
@@kelvinvu1247 No this is not true. If any, not all, of the three conditions are violated, it is not a context free language. What OP is saying is correct and is not the same as what you are saying. Essentially, these 3 conditions are necessary but not sufficient for proving a language is context free, they can only show you definitively that a language is NOT context free.
Thank you! Well explained!
1. Let L be the language {a^i b^i c^ / 0
thank you sir
Before i am struggling to pass now i am struggling to top 👍❤️
3:52 is it necessary to write this steps in my university examination sir ??
This is Amazing 😍
Thanks for the tutorial 👍👍
Nice explanation
thanks.well explained
Thank you
Can we make a parse tree in pumping lemma for Context free language.
amazing video.
Thankyou sir
Please put all the vedios we are going to have our semester exams
Thank you..
Thanks - Done
,./';
Thank you. Where is the playlist from this video?
goo.gl/f4CmJw
What if there are 3 ways for the string to be divided and one of ways shows you can pump it while the other two don't? Is the language context free?
if its pumpbable at least in one of the division possibilities then the contradiction proof fails. To understand this better you have to walk slowly through the thought process of the proof: We ASSUME that the language is pumpable. Therefore, it MUST HAVE a pumping length, call it P, so that ALL words longer than P MUST BE PUMPABLE. From here, we set to prove the contradiction: since our claim has led to saying ALL words longer than P are pumpable, it's enough to find One word that can't be pumped to prove the contradiction. If you picked a word and it turns out to be pumpable, you should try to find a different word that won't be; if you cant find one, it might be the case that the language is actually context free (it's not guaranteed, pumping lemma doesnt prove anything)
Thnku
why we divide her to 5 and what the intuition of that !?!
CFL closed under concatnation is valid but
L1={a^n, n>0}
L2={b^nc^n, n>0}
L1.L2={a^nb^nc^n} --> whic is not CFL
Then How it is closed under CFL?
Please help me........................
Well, instead of using the same index 'n' both times maybe using the index 'm' in L2 will make this more clear to you why the concatenation is still a CFL.
index : also called : Bar-Hillel lemma
TOMORROW IS MY EXAM
WEEEEEEE NEEEEEEEEEEED EXXXXXXXXXXAAAAAMMMMPLEEESSSSSSS
This channel does an example video, as a follow up to this video.
ua-cam.com/video/eQ0XkUk3qGk/v-deo.html
you need to get out of weed.
prove by example
Plz show an example....
mantul mantab betul
pasti mau kuis ya
Step 6 and 7 are in wrong order. Firstly should find ways of dividing as uvxyz, then consider some strings divided in that way not belong to A.
check abc
actually you should explain with an example
actually you should watch full video, he said in next lecture
SAVIOUR FR.
wawwwww
your voice is not good as I have to listen to it all the time.
so please give it a look
chutiya
WOULD it be good if u didn't have to listen to it all the time??you really make no sense....
What a douche
Just saw a few seconds of your Li Fi presentation, looked like a fucking fourth grader made it
He's from India...