Tmrw is my TOC exam. Wish me good luck. Sleep less nights. If you guys see this message pls do reply to this, I will see the comment which will later bring me nostalgic feels
@@aymanrahman4118 hello mate thanks for the reply I passed TOC and got S grade I completed my clg and now i am gonna work at Amadeus Bangalore as a SDE from coming January 🙂
I love how you quickely address questions that seem to be contradicting after you introduce a new concept, I be like, "wait! didn't you just say..." and 10 secs later I'll hear you say "Now you might be asking why did we do this when I said we can't". Awesome videos man. Been watching for tha past three months.
I did'nt understand turing machine before but, by watching your videos i have understood it very well. thanks so much. Your voice and background was clear. Over all this learning was awesome.
wow! this is much simpler then how my professor taught this in class! Made assignments so much easier! I'd rather be paying people like you these University fees than my professor...
Had my end sem exam Today, Started preparing yesterday, in 1 day covered almost all the topics and I think I'll score good as I solved all the questions. Thank u so much Neso, can't thank you enough!
I am studying CS at one of Australia's best colleges, and yet the cant teach as good as you and the few amazing youtubers out there. I believe this entire education system is flawed... I mean atleast you dont need to go to a college to study something like graphic designing or CS.... or rather something which requires techincal skill. Ofcourse colleges are important for medical sciences. back to the topic - brilliant video and great explanation.!
I read Peter Linz' book and still had some problem understanding the turing machine. Thank you so so much for explaining it this easy! It's 3am, I'm so sleepy and have my final exam at 11am, yet I totally got what you did. :D Thank youuu
in 10:45 , there is no need of separate D state right, we can loop through y -> y, R at A itself and the B -> B,L transition from A to acceptance state
The state D is required to prevent strings of the form (0^n)(1^n)(0^m)(1^m) from passing, dry run 0101 and you will see it will pass if you do not have the state D.
Thank you for the tutorial but if you can also explain "how to design the turing machine" that would be better instead of the "how works the designed turing machine".
Say i have a string "00010111" which is technically wrong according to {0^n 1^n | n >= 1}, but still the turing machine designed above will accept it right? how or i m making a mistake somewhere
This goes through A,B,B,B,C,C,C,A,B,B,B,B,C,C,C,C,A,B,B,B,B,C,C,C,C,A,D and gets rejected there for encountering 0 on the tape (which is not defined i.e. goes to reject).
1. So now the input is part of the tape instead of a separate input? 2. Can this algorithm be performed in a non-destructible manner? (that is, without destroying the original input on the tape by replacing it with Xs and Ys) 3. As defined so far, this Turing machine seems to be only capable of taking some input and accepting it or rejecting. But what about producing some OUTPUT? In order to perform some more useful computation, the computing machine should be able to calculate some results (e.g. evaluating arithmetic expressions) and output them. Can we do that within the current framework? 4. What about PROGRAMMING the machine? Isn't it that we have to make a special machine for every single tast we want to perform with this particular design? Because the program is stored in the "head", as the state automaton, and this automaton has to be replaced by another one in order to change how the machine works.
Brilliant!!! Neso Academy is the best, you are tool much team. Keep up the good work and maintain the same level of standard and even higher. You are great. Do you offer Online certificates? i need it
Why can’t if we encounter 1 then we move left continuously and encounter blank for accept state. Here we can save memory for not taking extra space for y. If it is double taped Turing machine.
Very excellent explanatory video. Thank you very much! I just have one question. What are the sets Σ and Γ for this particular Turing machine? I could guess that one of them could be {0, 1, x, y} but I don't understand their differences.
In your program, why didn't you just have the transition function δ(A, y) -> δ(accept, y, R)? I'm using "accept" there as the halt and accept state. Isn't that more efficient than going all the way to the right?
If we encounter 1 at the very first place what are we replacing it with... I mean I understood it will go to reject state but how will we represent it...
If it could, I think he would give us an example. Since a TM has a abilty to write over the current symbol, every each program cpuld be written this way, no matter what kind of program we are talking about. This is actually the easiest possible way of counting.
If you have a laguage that accepts the same number of 0's and 1's, you start with poping 1 or 0, then u push every 1 with adding 0 on stack and vice versa. If u get to the empty stack, that means every 1 that was popped on the stack was pushed by every 0, which means that u had the same numbers of 0 and 1, and the string was accepted.
Neso Academy----The Savior of Nights Before Exam ✌✌✌✌😂😂
Tomorrow is my toc exam 😭 this is lit
Tmrw is my TOC exam. Wish me good luck. Sleep less nights. If you guys see this message pls do reply to this, I will see the comment which will later bring me nostalgic feels
Literally 💯
@@10subsonlychallenge66 how did the exam go?
@@aymanrahman4118 hello mate thanks for the reply
I passed TOC and got S grade
I completed my clg and now i am gonna work at Amadeus Bangalore as a SDE from coming January 🙂
I love how you quickely address questions that seem to be contradicting after you introduce a new concept, I be like, "wait! didn't you just say..." and 10 secs later I'll hear you say "Now you might be asking why did we do this when I said we can't". Awesome videos man. Been watching for tha past three months.
I agree. To me, that's usually the mark of a great and talented teacher 👌
I can't imagine what the people in 2016 and earlier had to go through without this channel
literally helped me to cover the whole syllabus one night before the exam kudos to neso
did u pass?
@@miteshshetty6600 🤣😂😂
@@miteshshetty6600 🤣🤣🤣🤣🤣🤣🤣🤣
us bro us
I've my TOC exam in 2 hrs and here I am learning the concepts in the best way possible. Thanks Neso Academy!
how was the exam dude? c:
@@042_md.shabuktahaidereric8 I did it well brother!
@@mohammedafzal4343 Barakallah..
I did'nt understand the turing machine before but, by watching your videos i have understood it very well. thanks so much .
you're fantastic explainer
awesome work u guys,free lectures of this much excellence level is rare .thank you so much man.
NESO_Academic is the best youtube channel all doubts are cleared and explain everything about the topic. I like very much the way of teaching👍
I did'nt understand turing machine before but, by watching your videos i have understood it very well. thanks so much. Your voice and background was clear. Over all this learning was awesome.
The background of your video and clearity of your speech forces me to see your videos keep making it 👌
true
Thanks for filling in for my prof! I don't think people like you truly realize the level of help that you provide. Wish you all the best
Learned a lot more than being in the class. Thanks to Neso Academy..
This is a really good tutorial. Thank you for making a detailed walk-through of a TM!
Big love. These tutorials are all I need to pass my AT class fr. Tyvm
wow! this is much simpler then how my professor taught this in class! Made assignments so much easier! I'd rather be paying people like you these University fees than my professor...
Had my end sem exam Today, Started preparing yesterday, in 1 day covered almost all the topics and I think I'll score good as I solved all the questions. Thank u so much Neso, can't thank you enough!
Kaisa hua
@@shivaish4355 bhai out of 60 mein ne 54 score Kiya Tha.
@@boringmangesh tips bata bhai koi mera compartment ka exam h
You explained it like anyone who doesn't know the basic idea about Turing Machine can also understand and learn... Thank you
I am studying CS at one of Australia's best colleges, and yet the cant teach as good as you and the few amazing youtubers out there. I believe this entire education system is flawed... I mean atleast you dont need to go to a college to study something like graphic designing or CS.... or rather something which requires techincal skill. Ofcourse colleges are important for medical sciences.
back to the topic - brilliant video and great explanation.!
I got mind blown on this tutorial, It's amazing the way I understand each process, thank you for making this amazing tutorial
I read Peter Linz' book and still had some problem understanding the turing machine. Thank you so so much for explaining it this easy! It's 3am, I'm so sleepy and have my final exam at 11am, yet I totally got what you did. :D
Thank youuu
Thanks for making the life of thousands of students much easier. Respect!
I always feel thankful towards you whenever I watch your video. Kuddos to your hardwork.
One of the best available online tutorial
Every topic is very well defined and it proves helpfl for me during exams.
Thank you very much
You explained to me in 13 minutes what I couldn't understand in a 2 hours lecture in college
Great work sir
Always get satisfied when i study from ur lectures..
Amazing video sir.. I got each and every point you explained.. Your way of teaching is really understandable..
in 10:45 , there is no need of separate D state right,
we can loop through y -> y, R at A itself and the B -> B,L transition from A to acceptance state
The state D is required to prevent strings of the form (0^n)(1^n)(0^m)(1^m) from passing, dry run 0101 and you will see it will pass if you do not have the state D.
thank u so much, good video not only save grades, but it save fellow stem souls!
Love and respect from Bangladesh.
Neso academy =lifesaver
Great explanation everytime👍
3 am , night before finals.
Neso academy here i come again....
Thankyou ❤️
It's so helpful and easy to understand. 😊
thaks, can be repetitive, but you teach excellent and bluntly
is the accept state care about that where the head on the tape?
if not, why we should "blank->blank, L"?
can we use "blank->blank, R"?
I think it does'nt matter that you go left or right once you have completed your string it doesn't matter much
I think both are correct, its totally depends to u that whether u write b->b,L or b->b,R.
Thank you for the tutorial but if you can also explain "how to design the turing machine" that would be better instead of the "how works the designed turing machine".
جزاك الله خيرا. شرحك رائع و مبسط
Thank you so much for this awesome explanation :) easy to understand, keep helping us by making such nice videos :)
Say i have a string "00010111" which is technically wrong according to {0^n 1^n | n >= 1}, but still the turing machine designed above will accept it right? how or i m making a mistake somewhere
This goes through A,B,B,B,C,C,C,A,B,B,B,B,C,C,C,C,A,B,B,B,B,C,C,C,C,A,D and gets rejected there for encountering 0 on the tape (which is not defined i.e. goes to reject).
night before midterm, lets gooooo
Me too I just smoked some gelato 🍧 rip
@@trippingtonproductionsco.7641 Guess wut, Im here again for the final RIP good luck my friend
@@justinthenerd561 I feel u
@@justinthenerd561 what about now?
@@aydict I did it!
Great Explanation
U r just amazing.. Thank you 🌻
Thank u so much. ... u guys made it simple to understand
Really great Explanation.😮😮
Oh my god you explain it so well
Very nicely explained. Thankyou
Perhaps no one can explain it better
1. So now the input is part of the tape instead of a separate input?
2. Can this algorithm be performed in a non-destructible manner? (that is, without destroying the original input on the tape by replacing it with Xs and Ys)
3. As defined so far, this Turing machine seems to be only capable of taking some input and accepting it or rejecting. But what about producing some OUTPUT? In order to perform some more useful computation, the computing machine should be able to calculate some results (e.g. evaluating arithmetic expressions) and output them. Can we do that within the current framework?
4. What about PROGRAMMING the machine? Isn't it that we have to make a special machine for every single tast we want to perform with this particular design? Because the program is stored in the "head", as the state automaton, and this automaton has to be replaced by another one in order to change how the machine works.
very good viedo!Very helpful to understand , thank you a lot!
Sen nasil bi kralsin..
one night batting full marks with ur lectures thanks sir
Brilliant!!! Neso Academy is the best, you are tool much team. Keep up the good work and maintain the same level of standard and even higher. You are great. Do you offer Online certificates? i need it
This lecture is gold...tq sir
Soo nice to understand thanks sir for helping.
watching this right 4 hours before my final
can we put the transition for 'A' state for blank -> blank , R and goes to final state?
Why can’t if we encounter 1 then we move left continuously and encounter blank for accept state. Here we can save memory for not taking extra space for y. If it is double taped Turing machine.
You are Amazing!
bhai bhai neso is saviour
7:11 should we draw "reject" state ?
super clear! easy to understand
Very excellent explanatory video. Thank you very much!
I just have one question. What are the sets Σ and Γ for this particular Turing machine?
I could guess that one of them could be {0, 1, x, y} but I don't understand their differences.
Σ = {0, 1}
Γ = {x, y}
Because Σ is alphabet of the language L and Γ is a set of symbols in the tape
@@filipflajsman2027 Wrong. Σ and blank belong to Γ.
can we have a self loop on A for y - - - ->y , R rather than going into D state and having self loop there?
Great explanation
thx a lot sir
Thank you, for a great video.
In your program, why didn't you just have the transition function δ(A, y) -> δ(accept, y, R)? I'm using "accept" there as the halt and accept state. Isn't that more efficient than going all the way to the right?
so good, thank you very much !!!
thank you it help me so much ;)
If we encounter 1 at the very first place what are we replacing it with... I mean I understood it will go to reject state but how will we represent it...
thaks a lot sir ,very well explained
Thanks a lot sir ji😊👌
Does this work for 01 (eaching appearing once)? I don't think so. Flawed machine.
Is it compulsory to mention accept state or it can be excluded like reject state?
I'm so much satisfied.
how do we decide the direction just before ACCEPT?
How to make Transition Diagram for The Turing machine
what is the need of state D cant we just again again see Y and move right from A itself?
Very Helpful
I have one question sir, how can we can design a TM such that (0^n 10^3n) how do we solve this?
Thank you
nice lecture
Changing 0 to X and 1 to Y is a rule? Or can we change it
Wonderfull explanation
thank you very much sir
Why have you writyen transition using y at state B please explain?
You are incredible good
Can you give an example of a rejected string?
0001
Does it really matter that we go either left or right after reaching the final state?
it doesnt matter really
Thank you so much for Your videos
Thank u!!!!! Life saver
Greatful to u........ 😊
shouldnt 1->Y be move right and not left for B to C
You never told how you come up with the diagram?
Ok that's right
Can I define the algorithm for this example differently?
can a turing machine to count number of 0s and 1s be drawn? It seems to be simpler than this one. If yes, then how?
If it could, I think he would give us an example. Since a TM has a abilty to write over the current symbol, every each program cpuld be written this way, no matter what kind of program we are talking about. This is actually the easiest possible way of counting.
Can't we do the same problem with PDA with much lesser complexity?
Yes we can do
If you have a laguage that accepts the same number of 0's and 1's, you start with poping 1 or 0, then u push every 1 with adding 0 on stack and vice versa. If u get to the empty stack, that means every 1 that was popped on the stack was pushed by every 0, which means that u had the same numbers of 0 and 1, and the string was accepted.
Easy yes. They have got a tutorial for that too.
Thank you so much!!!!!
when the tape moves from left to right it must see the 1 and then replace it but in your turing machine you havent mention 1->y,L when you reach at A.