Is this approach, i.e., introducing epsilon between the inputs, inefficient ? Because for each input, you have to explore 2 possibilities. Is there way to do pruning so that we can stop the wrong branch as early as possible ?
Why didn't you take the case of epsilon on the first state when we are at q2, I think when we have an option like 'epsilon and then a letter,' then we can choose epsilon or the letter, but if we have the case like 'a letter and epsilon', we can only choose the letter and we don't have a choice.
A little question. It seems to me that this PDA will accept an empty string as well, because it can perfoem an epsilon transitions all the way to q4 in the following manner: 1. ε-transition from q1->q2, pop nothing, push z0, . 2. ε-transition from q2->q3, pop nothing, push nothing. 3. ε-transition from q3->q4, pop z0, push nothing. So I think this PDA may accept an empty string or even reach to q4 before any input. What am I missing?
The current position of this lecture in the playlist is #79, which is wrong! The placement should have been done after #88(current order). Please UPDATE the playlist order. Thank You.
sir i have watched all your videos from the begining and they have really helped me a lot in all aspect but sir i this some part of the video is missing before this video beacuse you have suddenly started push down automata from nowhere and i am getting bit confused
Order of this video is wrong. It could cause confusion among students since it is placed here.It should be placed after pushdown automata introduction and after part 1 and 2.
so I have one doubt regarding this explanation, see we have the string in form "eaebeaebe" where e stands for epsilon, right? so at the very first transition we can either get 1st e or a as an input, but as we have no Non-deter PDA we don't have any branch for input a on state q1, right?, so simple you have neglected first branch and took e as only the option as input, then second transition you have taken a which seems to be correct, till now we have inputted "ea" okay? now you have took 2 transition one for e (so the string form inputted after this transition "eae") and one for b (so the string form inputted after this transition "eab") , completely right! Nowwwwwwwwwwwwwwwwwwwwwww, after we have explored the branch "eae" why do we have two options as next transition , after "eae" we should have only one option "b" as an input and not "e", as after and only after input "b" it can have the access to the next epsilon (the third epsilon) !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! "eae" ->b = "eaeb" -> e|a
Correct way: (Tutorials Point) (bit wrong in Neso). At first, we have to assume that L is regular. So, the pumping lemma should hold for L. Use the pumping lemma to obtain a contradiction − Select w such that |w| ≥ c Select y such that |y| ≥ 1 Select x such that |xy| ≤ c Assign the remaining string to z. Select k such that the resulting string is not in L.
but isnt an empty string the same forwards and backwards? if E is my empty string (E = " "), then E == reverse(E) right? and E has length 0, which is even, so it should fit the even palindrome machine.
You, sir, have provided the best lecture "trio" I have ever watched.... lec - 88, 89, 90.
absolute masterclass content
triptych
What are you doing today
This was the perfect explanation i was looking for in part 1
Your videos are helping so much, but do you have an example of this with a transition table?
Brilliant explanation!
alignment is improper this video should be at 88 place of this series
after 90, PDA PART-2
That's not a big deal 😑
ah thank you, was so confused
Is this approach, i.e., introducing epsilon between the inputs, inefficient ? Because for each input, you have to explore 2 possibilities. Is there way to do pruning so that we can stop the wrong branch as early as possible ?
not a man but a legend
Why didn't you take the case of epsilon on the first state when we are at q2, I think when we have an option like 'epsilon and then a letter,' then we can choose epsilon or the letter, but if we have the case like 'a letter and epsilon', we can only choose the letter and we don't have a choice.
Sir, How much time will you take to upload complete course of TOC?
Please upload remaining videos I have exam next month, your videos helped me lot.
is pda and determininistic pda the same?? what about npda ?? equivalence of npda & cgf ... please upload this videos as soon as possible......
can i have the video of odd pallindrome? on urgent basis
great sir
A little question.
It seems to me that this PDA will accept an empty string as well, because it can perfoem an epsilon transitions all the way to q4 in the following manner:
1. ε-transition from q1->q2, pop nothing, push z0, .
2. ε-transition from q2->q3, pop nothing, push nothing.
3. ε-transition from q3->q4, pop z0, push nothing.
So I think this PDA may accept an empty string or even reach to q4 before any input.
What am I missing?
Technically empty string may be considered as an even palindrome
i think they assume empty string to be an even palindrome
Incorrect. Due to there only being a single epsilon, as you described, the PDA would get stuck at q2, causing it to be unacceptable.
The current position of this lecture in the playlist is #79, which is wrong!
The placement should have been done after #88(current order).
Please UPDATE the playlist order.
Thank You.
Thanks for mentioning which is helpful now
how it will work for abbbba??? because at the second 'b' there would also be 'b' in the stack at top most.
Plz add next lecturers plzz...
thank you so much
12:35
sir i have watched all your videos from the begining and they have really helped me a lot in all aspect but sir i this some part of the video is missing before this video beacuse you have suddenly started push down automata from nowhere and i am getting bit confused
this video's correct position is at 91,after 90-> PDA(Even pallindrome part2),so skip it for now
Thankyou sir
can you give me the link where you've discussed the examples 1&2 for push down automata
sir shall we write the state transition diagram only in the exam or we have to show the the stack diagram also
One little thing : This video isnt in the right place of the Playlist, beside that its a good video
Please, someone explain if two states are active simultaneously , which one access stack content first?
NPDA is a hypothetical model and can't be implemented practically.
You have misplaced the videos 87 and 78. please do the needful. :)
Where are the part 1&2 for pushdown automata
How come he doesn't include the epsilon right in the beginning? Because it can be eeeaebe...
how to design pda ??
I can't find Turing machine explanation??
Order of this video is wrong. It could cause confusion among students since it is placed here.It should be placed after pushdown automata introduction and after part 1 and 2.
so I have one doubt regarding this explanation, see we have the string in form "eaebeaebe" where e stands for epsilon, right? so at the very first transition we can either get 1st e or a as an input, but as we have no Non-deter PDA we don't have any branch for input a on state q1, right?, so simple you have neglected first branch and took e as only the option as input, then second transition you have taken a which seems to be correct, till now we have inputted "ea" okay? now you have took 2 transition one for e (so the string form inputted after this transition "eae") and one for b (so the string form inputted after this transition "eab") , completely right!
Nowwwwwwwwwwwwwwwwwwwwwww, after we have explored the branch "eae" why do we have two options as next transition , after "eae" we should have only one option "b" as an input and not "e", as after and only after input "b" it can have the access to the next epsilon (the third epsilon) !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! "eae" ->b = "eaeb" -> e|a
there can be any number of epsilons before and after a character
If we reached the final state but our stack is not empty then the string is accepted or not
accepted
how pda decide string is valid or not
Will palindrome PDA fail or not for abbabb
Like it is not a palindrome but the PDA accepts it .Or does it not?
It is not a palindrome. But do you mean like
A push
B push
B pop
A pop
B push
B pop ??
Correct way: (Tutorials Point)
(bit wrong in Neso).
At first, we have to assume that L is regular.
So, the pumping lemma should hold for L.
Use the pumping lemma to obtain a contradiction −
Select w such that |w| ≥ c
Select y such that |y| ≥ 1
Select x such that |xy| ≤ c
Assign the remaining string to z.
Select k such that the resulting string is not in L.
What the heck is that? what is the difference between part 2 and 3 omg
it seems a string of 3 consecutive epsilon, which still is a empty string, can be accepted, but it shouldn't in this case
but isnt an empty string the same forwards and backwards? if E is my empty string (E = " "), then E == reverse(E) right? and E has length 0, which is even, so it should fit the even palindrome machine.