Youre a fantatic teacher. My professor cant teach this in a 2 hour lecture and yet you did it so clearly in 9 minutes and 14 seconds. Why do I go to class?
My professor did a similar example today in class as an introductory lesson to PDA, and I had no clue what she was talking about. I immediately understood this video. Thanks.
Great example and works for every CFG. The only thing I would add is if you have recursion in the same production like S --> abSb | epsilon then we have to make another loop to remove S from the stack within the loop state mentioned in the video.
So what if instead of B -> epsilon as a production in our CFG, we had B -> a (or B-> b). Could we still use a self loop back to qloop. In other words can we use a self loop to qloop, anytime the RHS of the production has only one symbol? Excellent video btw!
Hi there, could you remove the pop-ups at the end of the video that goes to another video because it blocks your writing and we could not see anything.
Shouldn't the transition from the second state to q-loop be (epsilon, $ -> S) .. since the input read is nothing, stack top is a dollar and to push is S?
Why would you pop $ there? The whole purpose of putting the $ on the stack at the very start is to ensure we can't go to the final state unless the stack is "empty" (other than $).
@@DiwashHCR2 This is where divergence in notation comes into play -- my notation (based on the Sipser book) is: (input_symbol, thing_to_pop (or not), thing_to_push (or not)). So the first transition here pushes a $, and then pushes an S (the start variable). So when we first enter q_loop, the stack contents are $S (top of stack on the right).
There are some books that do force something to be popped on each transition; some others (maybe yours) does a "peek" at the top but doesn't pop; others can (maybe yours also) forces a push-only or pop-only transition.
Thanks for the video, i have a question, i don't understand when you're in qloop why isn't it S, S ; c instead of epsilon, S ; c would make more sens for me if it was S,S ; c no ? Or does both work. Thanks
Great video, but why not accept with empty stack and simply use two states: (We have to assume that $ is already on the stack, though. Otherwise, this automaton would accept every input.) The first state pushes the start-non-terminal on the stack and the second state loops over itself, while using every single rule from the CFG and pops terminals from the stack, while pushing nothing on the stack. The moment the stack is empty again, simply go back to the first state and push nothing on the stack. Our automaton should accept now, because the stack is empty.
Or, if we want to have an accepting state, we could add a third state and if we have an empty stack (even without $), then we go from the second state to the third one and accept.
You are the best Automata Theory youtube channel. It's a niche title, but you earned it.
I'll accept my award and parade soon :) (thanks very much!)
I agree
Youre a fantatic teacher. My professor cant teach this in a 2 hour lecture and yet you did it so clearly in 9 minutes and 14 seconds. Why do I go to class?
Finally, a short and to-the-point video clearing out the concepts! Thanks ✌🏻
You're welcome!
@@EasyTheory yes please do more vids like this, it's more efficient and easy to understand.
@@EasyTheory is that possible to do a same version for pda to cfg?
My professor did a similar example today in class as an introductory lesson to PDA, and I had no clue what she was talking about. I immediately understood this video. Thanks.
@@forthehomies7043 Ur professor = Trump
Easy Theory = Kamala
Great example and works for every CFG. The only thing I would add is if you have recursion in the same production like S --> abSb | epsilon then we have to make another loop to remove S from the stack within the loop state mentioned in the video.
This is the best explanation so far. It is concise and to the point. Thanks for this video.
You're very welcome!
I have an assignment due in an hour and you may have just saved me a letter grade. Thank you!
The man, the myth, and the legend of theory of computation and teaching in general!
You are THE MAN. Thanks for this awesome explanation!
Alvi Habib thanks very much! Make sure to check out the lecture series I'm currently doing.
FINALLY an English Video thank you so much man this was great
Lol thanks!
wow. thank you so much. I was watching other videos so confused by all the math. your video made it super simple, thank you.
This is the best PDA explanation video that I've found, thank you!
i luv the clear way u describe. THX for saving my final💪
clear, concise, and comprehensive.
you are a godsend, thank you
sir. you've saved my life
you saved me, way better explanation than my professor
You're welcome!
You're the man I take refuge in 😢🙌🏻💖
Amazing video Sir Alan
Super helpful! I understand it so much better now.
Thank you!!! This made so much more sense than the other explanations i found
LMFAOOOOOOOOOOOOO thank you man this is literally what i needed
Man, It was 6th video in my RUclips search "cfg to pda" which I understood.
Thanks man.
5 more spots to rise up! ;)
@@EasyTheory Yeah, The volume was a bit slow, but headphones worked fine for me.
Perfectly explained, bless you 🙏
omg you are literally the best
So what if instead of B -> epsilon as a production in our CFG, we had B -> a (or B-> b). Could we still use a self loop back to qloop. In other words can we use a self loop to qloop, anytime the RHS of the production has only one symbol?
Excellent video btw!
To answer your question a year later, yes you should be able to simply use a self loop back to qloop when the RHS of the production has one symbol.
Fantastic video! This helped me so much!!!
This was such a great video! Thank you so much!
Thank you so much, you saved a lot of lives!!
You're very welcome!
Good explanation! Much appreciated
So how do you represent S -> A, would this just be a self loop on qloop being (epsilon, S -> A)?
Yes, correct! Or you can have two transitions that "go out of qloop and come back" but that's not necessary.
Thank you!! This is a great video
Thank you so much now is easy to do my homework.
Thank you thank you thank you so so so much!!!! I subscribed
You're very welcome!
Thanks for this! Finally, i got it ✨
Great!
How would this look if we accept by final state rather than empty stack?
Thank you so much, really helped!
Another great vid 👌🏻
What if the language has no terminals? Then what would go in qloop
Thanks for neat explanation
Hi there, could you remove the pop-ups at the end of the video that goes to another video because it blocks your writing and we could not see anything.
Great Video !
Thank you so much man
Shouldn't the transition from the second state to q-loop be (epsilon, $ -> S) .. since the input read is nothing, stack top is a dollar and to push is S?
Why would you pop $ there? The whole purpose of putting the $ on the stack at the very start is to ensure we can't go to the final state unless the stack is "empty" (other than $).
@@EasyTheory Thanks ... I thought the tuple was (input symbol, stack top symbol, push/pop)
@@DiwashHCR2 This is where divergence in notation comes into play -- my notation (based on the Sipser book) is: (input_symbol, thing_to_pop (or not), thing_to_push (or not)). So the first transition here pushes a $, and then pushes an S (the start variable). So when we first enter q_loop, the stack contents are $S (top of stack on the right).
There are some books that do force something to be popped on each transition; some others (maybe yours) does a "peek" at the top but doesn't pop; others can (maybe yours also) forces a push-only or pop-only transition.
What is the software you use for drawing? It is beautiful.
did you know what it is?
Hi there, I was wondering if you have any videos where we can go backwards? Going from a PDA to a cfg
Video coming out soon :) the CFL livestream happening in a few weeks will certainly cover this too
Thanks for the video, i have a question, i don't understand when you're in qloop why isn't it S, S ; c instead of epsilon, S ; c would make more sens for me if it was S,S ; c no ? Or does both work. Thanks
Because S, S -> c means "read an S" but that is already a problem because the input is over the input alphabet Sigma, not the stack alphabet Gamma.
Thank you sooooo sooooo much!!!!
Thank you so much!
i love you.
what is the significance of the read only self loop?
is that the case even if a terminal isnt in the S(start) rule?
Thank you!!
Loved It
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Thanks a lot!!!!
You're welcome :)
Thanks!
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Great video, but why not accept with empty stack and simply use two states:
(We have to assume that $ is already on the stack, though. Otherwise, this automaton would accept every input.)
The first state pushes the start-non-terminal on the stack and the second state loops over itself, while using every single rule from the CFG and pops terminals from the stack, while pushing nothing on the stack. The moment the stack is empty again, simply go back to the first state and push nothing on the stack. Our automaton should accept now, because the stack is empty.
Or, if we want to have an accepting state, we could add a third state and if we have an empty stack (even without $), then we go from the second state to the third one and accept.
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I used this tutorial for a quiz and my professor marked it completely wrong :(
did he say why?
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Why we can't take all these in single loop on qloop state ?
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what
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