Context-Free Grammar to Pushdown Automaton Conversion (CFG to PDA)

[@goobles3991]

You are the best Automata Theory youtube channel. It's a niche title, but you earned it.

[@EasyTheory]

I'll accept my award and parade soon :) (thanks very much!)

[@MUSHIN_888]

I agree

[@prakhartiiwarii]

Finally, a short and to-the-point video clearing out the concepts! Thanks ✌🏻

[@EasyTheory]

You're welcome!

[@Shan-gn7mg]

@@EasyTheory yes please do more vids like this, it's more efficient and easy to understand.

[@Shan-gn7mg]

@@EasyTheory is that possible to do a same version for pda to cfg?

[@sebastianllano1405]

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?

[@forthehomies7043]

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.

[@inarazahin5786]

This is the best explanation so far. It is concise and to the point. Thanks for this video.

[@EasyTheory]

You're very welcome!

[@thatguy7802]

I have an assignment due in an hour and you may have just saved me a letter grade. Thank you!

[@konstantinsazhnev]

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.

[@alvihabib]

You are THE MAN. Thanks for this awesome explanation!

[@EasyTheory]

Alvi Habib thanks very much! Make sure to check out the lecture series I'm currently doing.

[@TheWiimaster456]

FINALLY an English Video thank you so much man this was great

[@EasyTheory]

Lol thanks!

[@techademy9354]

The man, the myth, and the legend of theory of computation and teaching in general!

[@matthewsocoollike]

wow. thank you so much. I was watching other videos so confused by all the math. your video made it super simple, thank you.

[@ashleyarmenta2289]

This is the best PDA explanation video that I've found, thank you!

[@Leonidesu]

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.

[@沼氨]

i luv the clear way u describe. THX for saving my final💪

[@Dakun]

you saved me, way better explanation than my professor

[@EasyTheory]

You're welcome!

[@zeitgeist18]

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!

[@moatef1886]

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.

[@wengeance8962]

So how do you represent S -> A, would this just be a self loop on qloop being (epsilon, S -> A)?

[@EasyTheory]

Yes, correct! Or you can have two transitions that "go out of qloop and come back" but that's not necessary.

[@ShaidaMuhammad]

Man, It was 6th video in my KZbin search "cfg to pda" which I understood. Thanks man.

[@EasyTheory]

5 more spots to rise up! ;)

[@ShaidaMuhammad]

@@EasyTheory Yeah, The volume was a bit slow, but headphones worked fine for me.

[@pk-le5bb]

clear, concise, and comprehensive. you are a godsend, thank you

[@DiwashHCR2]

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?

[@EasyTheory]

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]

@@EasyTheory Thanks ... I thought the tuple was (input symbol, stack top symbol, push/pop)

[@EasyTheory]

@@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).

[@EasyTheory]

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.

[@jaeger_404]

Thank you!!! This made so much more sense than the other explanations i found

[@shyamkannan8594]

What if the language has no terminals? Then what would go in qloop

[@dan-cj1rr]

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

[@EasyTheory]

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.

[@globalxoxo]

Amazing video Sir Alan

[@cadencetennant3303]

Super helpful! I understand it so much better now.

[@rosenv.1953]

How would this look if we accept by final state rather than empty stack?

[@albanyrebelion]

what is the significance of the read only self loop?

[@albanyrebelion]

is that the case even if a terminal isnt in the S(start) rule?

[@farwahbatool6247]

You're the man I take refuge in 😢🙌🏻💖

[@woodfur00]

Perfectly explained, bless you 🙏

[@pauldumitriu9237]

Hi there, I was wondering if you have any videos where we can go backwards? Going from a PDA to a cfg

[@EasyTheory]

Video coming out soon :) the CFL livestream happening in a few weeks will certainly cover this too

[@sonjamariemusic]

sir. you've saved my life

[@Patrick-S3]

Fantastic video! This helped me so much!!!

[@dilaragoral2259]

Thank you so much, you saved a lot of lives!!

[@EasyTheory]

You're very welcome!

[@melissamarlen1022]

This was such a great video! Thank you so much!

[@Jessislegend]

Good explanation! Much appreciated

[@KhangNguyen-wj5jd]

What is the software you use for drawing? It is beautiful.

[@fadyaldhaim4766]

did you know what it is?

[@rotisosis8285]

Thank you so much now is easy to do my homework.

[@cobyjonah7675]

omg you are literally the best

[@amanxo1]

LMFAOOOOOOOOOOOOO thank you man this is literally what i needed

[@FernandaRodriguez-xf3lr]

Thank you!! This is a great video

[@ananayarora]

Thank you thank you thank you so so so much!!!! I subscribed

[@EasyTheory]

You're very welcome!

[@cihadozcan1035]

Thanks for neat explanation

[@alexoxley5058]

Thank you so much, really helped!

[@evaeunicemoreno6349]

Thanks for this! Finally, i got it ✨

[@EasyTheory]

Great!

[@LPNorway]

Thank you so much man

[@justinkuang9423]

Another great vid 👌🏻

[@csperi-peri2447]

Great Video !

[@amynguyen3131]

Thank you so much!

[@jeffchong3450]

Thank you!!

[@zeruihai6051]

Thank you sooooo sooooo much!!!!

[@samjudelson]

Thanks!

[@nabeelbaghoor1061]

Thanks a lot!!!!

[@EasyTheory]

You're welcome :)

[@LasradoRohan]

Loved It

[@unpluggability]

i love you.

[@armanyama3186]

Why we can't take all these in single loop on qloop state ?

[@fatimaalamour5256]

In Germany we say Ehrenmann

[@princeelliot2836]

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.

[@princeelliot2836]

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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[@shaysarn2235]

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[@darklord998]

I used this tutorial for a quiz and my professor marked it completely wrong :(

[@beans_potatoes]

did he say why?

[@HornbyHardy-b2i]

Osinski Path

[@twondai2642]

what

[@fernandogmail1]

thank you 😭😭😭😭