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

@EasyTheory

3 жыл бұрын

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

@moeyali123

Жыл бұрын

I agree

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

@EasyTheory

3 жыл бұрын

You're welcome!

@Shan-gn7mg

Жыл бұрын

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

@Shan-gn7mg

Жыл бұрын

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

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.

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

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

@EasyTheory

3 жыл бұрын

You're very welcome!

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

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

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

You are THE MAN. Thanks for this awesome explanation!

@EasyTheory

3 жыл бұрын

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

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

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

Super helpful! I understand it so much better now.

Fantastic video! This helped me so much!!!

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

Good explanation! Much appreciated

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

@EasyTheory

3 жыл бұрын

Lol thanks!

Perfectly explained, bless you 🙏

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

@EasyTheory

3 жыл бұрын

5 more spots to rise up! ;)

@ShaidaMuhammad

3 жыл бұрын

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

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

You're the man I take refuge in 😢🙌🏻💖

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

@EasyTheory

3 жыл бұрын

You're very welcome!

Thank you so much, really helped!

Thank you!! This is a great video

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

Amazing video Sir Alan

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

@EasyTheory

3 жыл бұрын

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

Thanks for neat explanation

Another great vid 👌🏻

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

3 жыл бұрын

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.

Thanks for this! Finally, i got it ✨

@EasyTheory

3 жыл бұрын

Great!

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

2 жыл бұрын

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.

Thank you so much man

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

@EasyTheory

3 жыл бұрын

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

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

LMFAOOOOOOOOOOOOO thank you man this is literally what i needed

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

Thank you so much!

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

@EasyTheory

3 жыл бұрын

You're very welcome!

sir. you've saved my life

omg you are literally the best

Thank you sooooo sooooo much!!!!

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 !

you saved me, way better explanation than my professor

@EasyTheory

3 жыл бұрын

You're welcome!

Thank you!!

thank you 😭😭😭😭

Loved It

what is the significance of the read only self loop?

@albanyrebelion

23 күн бұрын

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

Thanks!

Thanks a lot!!!!

@EasyTheory

3 жыл бұрын

You're welcome :)

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

2 жыл бұрын

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

2 жыл бұрын

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

@EasyTheory

2 жыл бұрын

@@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

2 жыл бұрын

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.

i love you.

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

❤️

In Germany we say Ehrenmann

8:48

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

@beans_potatoes

3 ай бұрын

did he say why?

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

Жыл бұрын

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.

what