Lec-52: Design PDA for {w | na(w) = nb(w)} CFL language | Pushdown automata | TOC

[@ugcnet8015]

Sir, you deserve "Best Teacher Of The Year" award this time.

[@sagarm17]

Varun Sir are next level(totally different from other teachers) Computer Science Professor/Teacher in India ❤🎉😊. I can say that Varun Sir are the God given gift to all CS/IT Students to gain lot of Technical Knowledge. 😊 Be healthy sir ❤

[@gunjanchoudhary4632]

You are too good sir. I am preparing for UGC-NET 2020 and your videos are really helpful.

[@zenitsu4139]

clear hua ?

[@gunjanchoudhary4632]

Yes

[@zenitsu4139]

@@gunjanchoudhary4632 konsa college ma ha?

[@somyachouhan7108]

​@@gunjanchoudhary4632wow thodisi light dedo kitna padha, kitni der padha, kaha se padha, kaise padha, aur aap abhi kaha ho I mean kis field mai aur kya kr rhe ho

[@gunjanchoudhary4632]

@@somyachouhan7108 gate smashers and knowledge gate se padha mene, clear kr liya, paper 1 ke liye alg alg site h but usme b ache marks mil gaye the, currently government job h Rajasthan state me

[@jashandeep5165]

Sir are next level(totally different from other teachers) Computer Science Professor/Teacher in India ❤🎉😊. I can say that Varun Sir are the God given gift to all CS/IT Students to gain lot of Technical Knowledge. 😊 Be healthy s

[@safxxrahmed]

Sir you are a 'LEGEND' , and i mean it.

[@AnjaliSingh-uh3cl]

Please make the video fastly as we have to give semester exam you are saviour for us and great teacher you are lots of respect

[@imankit.paudel]

how does your semester go

[@biplobmanna]

The final transition rules should be somewhat like this: Let, q0 be the "initial" & q1 be the "final" state. (q0, 0, z0) -> (q0, 0.z0) (q0, 0, 0) -> (q0, 0.0) (q0, 1, 0) -> (q0, Ɛ) (q0, 1, z0) -> (q0, 1.z0) (q0, 1, 1) -> (q0, 1.1) (q0, 0, 1) -> (q0, Ɛ) (q0, Ɛ, z0) -> (q1, z0)

[@tanujgupta8576]

Bhai yaha pe state change nhi hogi kya kyuki hm 0 se 1 p gye?

[@Quester2023-xp7rb]

yes at last epsilon,z0 / z0 output is z0 itself sir wrote epsilon i.e epsilon , z0 /epsilon

[@sidrakanwal8151]

Happy teachers dayy ..only just because of u me and my friends a able to pass ourr examm..

[@avrajitkundu7179]

Tomorrow is my TOC exam and it is only because of you that I am getting confidence on this subject😊😊

[@gamersrinu2166]

bhai west bengal se hai kya ?

[@avrajitkundu7179]

@gamersrinu2166 Haan kaise pata chala?

[@csebangla3618]

​@@avrajitkundu7179 Tera title dekh k pata chala...

[@gamersrinu2166]

@@avrajitkundu7179 naam aur title dekh ke

[@animationcrust1993]

Haven't seen sir like big bro yet 🙌🙌 Great sir ❤️😊

[@TheAsimjan]

Well explained... Your tutorials are always awesome... Love from Pakistan ❤️

[@teksomegeek2062]

Sir, I think that there will be 3 more transitions, one for when 1 enters and the stack is empty i.e. (1,z/1,z), one for when 0 comes after 1 and pops the 1 in stack i.e. (0,1/epsilon) and one when 1 comes after 1 which is in the stack i.e. (1,1/11).

[@csesubjectwise6574]

kzbin.info/aero/PLyeuwglfy-Cuz4I6-_8PA6HPQ0weFmFgC

[@csesubjectwise6574]

Clear your doubts here

[@atulkrishnan4673]

yeah in total 5 states (2 final states)

[@mdme2891]

@@csesubjectwise6574 thank u sir u r vdo cleared my doubt

[@mdme2891]

kzbin.info/www/bejne/Y5LJe3R5dq1seNU this vdo will clear the doubt of y to change state and when

[@richasalan4870]

Sir why states does not change when we take 1 as a input ??? In other videos we change our states also when we change input ...

[@GamingUltra421]

because each 1 for each 0

[@absingh773]

Same thing i cant understand

[@HirakModak]

Me also

[@souvikpratihar1]

my doubt also

[@atishsinhaa]

Same doubt

[@harinath_mishra]

Sir thanks for uploading toc lectures......😊

[@sinanmohammadnahian4768]

You are the real life spiderman to me 💖

[@rvr61]

Sir, I think, if we take a language which starts with 1 e.g. 1001, then in this case your PDA cannot POP the first 1 as initially z0 was in the stack.

[@jacksonsunny1261]

Exactly

[@atulkrishnan4673]

yeah but using his logic we can complete the solution...final solution will have 5 states( 2 final states).

[@ashwiniyer454]

Actually in such a case, you can change your stack alphabet. Like for 1001. Push 1. Read 0 and Pop. Push 0. Read 1 and Pop

[@kg3217]

​@@atulkrishnan4673 2 States me ho sakta hai, `including` 1 final state. Bas saare combinations ko q0 pe point kardenge, like when a is top, input a or b, when b on top, input a,b; When Z0 on top, a and b as input so total 6 loop to self i think, then the final epsilon, Z0 / Z0 to final state

[@anirbanchand4196]

@@ashwiniyer454 Exactly. He has written the condition 1,0/e two times. Instead of that, we can add one more transition 0,1/e on Q0 ... I think this should work properly...

[@DeltaTech-fw3cd]

amazing treaching sir ji

[@DeltaTech-fw3cd]

Thanks 🙏

[@mohdmaazazhar885]

I have one doubt As the language have strings with equal number of 0's and 1's then 1100 should also be accepted but also there is no transition when we are at q0 and input symbol is 1 and stack has z0 as top eventually 1100 will not be accepted by given PDA

[@raushanray1154]

i think these three transition states are missing (1,1/11) (1,z/1z) (0,1/E)..........

[@vinitakeer515]

Waiting for next Lecture 🙏🙏🙏🙏🙏

[@csesubjectwise6574]

kzbin.info/aero/PLyeuwglfy-Cuz4I6-_8PA6HPQ0weFmFgC

[@csesubjectwise6574]

Get your syllabus here

[@nishasharma-sn3nq]

I m fully satisfy sir ...sir mcq ki video v upload kar do...practices k leyea ..thanku

[@continnum_radhe-radhe]

❤❤❤

[@VansheetaSahuCI]

Aap yellow hoodie me bht acche lagte ho , ur winter collection is so awwsmm

[@souravbhagat1358]

Sir computer graphics pr bhi vidoes uplaod kijye and thanks for this video

[@vinitakeer515]

Waiting for next video sir

[@engineerkaushal2982]

Thank you sir 🙏🏻

[@akm3168]

I have a doubt why did we not go on new state ?

[@gyanprakashverma8123]

Sir please aap iss series ko jaldi complete kr dijiye

[@tusharsahu8587]

Sir is PDA mein smjh nhi aaya ki aapne 2 baar (1, 0/E) kyon transition show kiya jabki hum same state pr hi they please explain kr dijiye!

[@kg3217]

Reason pata chala ?

[@tusharsahu8587]

@@kg3217 nhi yaar

[@sanjeevbengeri9193]

@@tusharsahu8587 that was by mistake kyuki wahaa unhone dono input keliye pop kar rahe hai toh nahi likha hai sochke likh diya hoga

[@GaganTyagi2000]

chal kya rha h pda me ..................................itni confusing to inception bhi nhi thi........................

[@harshitpal5558]

And what if the string starts with 1

[@manishachanda8089]

sir apne last me mistake kiye hn shayad,, when we r reaching the final state the transition should be {,z0/0 but apne { likhe hn thats mean that z0 is popped out

[@kanchanmalethia5012]

Sir dssb k liye class start krdo plz u r great teacher ..

[@mohammadfaisal3649]

Sir, thanks for these lectures....... i have a question : the two transitions you made are exactly same..... i,e 1,0/epsilon.... DO we really need to write this two times or just one time.????

[@sohamkumbhar21]

No, you dont have to write 1,0|ε two times as it is already written in loop but instead you should add one case 0,1|ε i.e if 0 arrives and there is 1 then we should pop.

[@rahuldevanshu]

@@sohamkumbhar21 first one is input and second one is top of the stack, so from your case 0 is input (which can be there) but I don't think 1 should be there, because we're inserting only 0 here so how top of the stack is 1?

[@PookiPoo]

@@rahuldevanshu for this case add 1,0 in input table after 010011 to get the desirable answer

[@sohail0474]

No but you have to pop out in the stack

[@komalprasadsahu6617]

@@sohail0474 do you also have exams tomorrow

[@aniketrajapure8251]

Sir in this sum for final state u wrote epsilon , zo / epsilon and in last video u wrote epsilon , zo / zo ..pls explain which to use when ?

[@educationwithhassannaseem2803]

love from pakistan sir

[@anmolpande7658]

can you please tell when to change the state and when not? I am a bit confused.

[@ankitadas1613]

Sir when we have to change the state??...in previous video we learnt that ,when new input come at q0, we should change the state...but in this case why it doesn't happen sir?? Can you please explain sir!!!

[@sohaibshamsi]

In that video we had to compare one 0 with two 1s but in this grammar we were given equal counts for 0s and 1s

[@833_sourinmukherjee7]

10:53 , I think it will be ε,z0/z0 . I dont know , I am right or wrong.

[@sambhalive6517]

You are right 👍

[@Nishantjain172k04]

No , at last step when there is no input in the input tape i.e epsilon and stack contains Zo then you have to pop out the Zo to make the stack empty Hence it will be €,Zo/€

[@Nishantjain172k04]

If you will perform €,Zo/Zo It means that you are simply ignoring the Zo and it is still left inside the stack

[@uwuwuw9237]

i wish youtube had a super-like button

[@shreyashchoudhary4576]

Very Helpful! Thanks a lot!

[@dh.418]

i have a doubt, in the previous example you changed state for every 1, but here there is a loop on only 1 state, pls explain

[@SALCEvaibhavadesara]

sir Last Transition should be ( E,z0/z0 ) but you mentioned ( E , z0/ E). sir Am i Right ? plz tell me ... Thankyou Sir , 👍

[@Patitapaban_sahoo]

Both are same you can use both ✌️

[@SALCEvaibhavadesara]

@@Patitapaban_sahoo Ok , Thanks 👍🏻

[@adityadhal5027]

thanks for asking this. I was also confused here.❣

[@novicemakers_abhishekkamal]

Helpful to me👍

[@asimbera1237]

when string starting with 1, we need to create another state q1 from q0 then do the same transition on q1 as q0 by replacing 0 to 1 and 1 to 0......

[@aashishkharal1444]

Yes! Thank you

[@nitishchaulagai3227]

Very underrated comment, everybody's doubt should be cleared after reading this

[@bhavyanayak8219]

Underrated

[@sambhalive6517]

Sir 10:53 pe yaha par €,Z0 | Z0 hona chahiye na 🙄, epsilon to empty stack pe lgta hai nai 👀 ?

[@gamersrinu2166]

haan bro

[@simransingh6137]

Thank you sir ❤️

[@shivanijain402]

Sir plz TOC ko continue karo...

[@GateSmashers]

Sure.. I will complete each and every lecture of toc soon

[@shivanijain402]

@@GateSmashers thank you sir

[@tushardixit5867]

Note: pda designed is not complete you have to make (1,0/z0) , (0,1/epsilon) and (1,1/11) states to q0 itself so that it can take those strings like 1100, 10..... and all those strings atq.

[@SonalDhani]

thanks u

[@trinmoydutta4879]

what if the input string is 011100?? after pushing first 0, 3 0's is to be popped...what to do then?

[@_ano_nym_ous_]

State kb change krte hain???

[@ankitachakraborty6909]

If string is - 011001 then it won't get accepted by the pda but here also we have equal no. Of a & b

[@gauravbhosle281]

True my friend. And also if... "1" comes in initial state "q0" and there is no "0" present ni stack then also it will fail... Basically...... Whenever we have 1 and there is no 0 present in stack then it will fail. Although we have equal no of zeros in our string but more number of 1 came first and PDA fails.

[@mexagrontoo]

@@gauravbhosle281 then 1 goes into the stack and the next zero will pop it out maybe?

[@SALCEvaibhavadesara]

@@gauravbhosle281 then we have to perform 1 as a PUSH & 0 as a POP operation...

[@sakshighosh9861]

Thnk you sir

[@dhirajdyandyan4504]

Notification ❌naughty fication ✔️

[@FACTFUSION-l5x]

Dtu?

[@mayurdhakite4815]

What will happen if the string 001110 or 00111100 is given for the same language, you have explained in the above video (i.e, n0(w) = n1(w) )?? How the stack will perform PUSH & POP operation?

[@keerthichakrabattula84]

push 1st 2 zero's and pop when for 1st two 1's again for 3rd 1 push it and when zero pop it.

[@keerthichakrabattula84]

Hope I made it clear. Although it's been 8 months since you asked this I hope this can help some one.

[@mexagrontoo]

@@keerthichakrabattula84 how to know how many states will be there? like in a^n b^2n there were 4 states (q0,q1,q2,q3) but in this one its only one state (q0). how do you decide that?

[@SALCEvaibhavadesara]

@@mexagrontoo Because we did not perform any SKIP operation . we don't need to skip any element. when we do skip operation then we have to change State.

[@absingh773]

Why we dont change the state when 1 arrives as done in previous vedio

[@parth_shah_99]

011100 what will happen in this case. We will not be able to pop 0s because there are 3 consecutive 1s.?

[@ayushjha1308]

Bhai order nhi dekhna hai no. Of 0s and no. Of 1s equal hona caahiye bus (011100) isme 0-3 hai 1-3 hai isliye accept hogi

[@supreetmavintop1829]

@@ayushjha1308 lekin sir ka explain kiya hua pda isko accept nahi karega kyu ki 1,Zo / 1Zo nahi lika hai

[@CodingSikhoBhojpuriMe]

2 lakh views ho gaye sir

[@ritikshrivastava9442]

5:55 sir TOS ko bahar nikal ke phele tos ko then input symbol agar sahi to dalna hai to koi faida hi nhi bahar nikalne ka to kiu consider kare is step ko

[@tushargupta5805]

looks like sir you messed up a little bit here you write {1,0 / E} two times and you forget to write {1,1/11}

[@PrinceKumar-0701]

Sir, ye last question me jo PDA bnaye hain. Usme 1 se start hone wale string accept kaise honge? For example 111000.... Iske liye alag se state bnane parenge na? Reply kr dijiyega sir.... Thodi confusion ho rhi hai.

[@suryapandey3905]

1,z0/1z0 , 0,z0/ 0z0 , 1,0/ e , 0,1 / e , 1/1, 11 , 0,0/ 00 . 6 self loop and one change of state. $/z0/z0

[@supreetmavintop1829]

@@suryapandey3905 exactly thnx for Clearing my doubt

[@SKG12]

It will not accept 101100 string It is partially right

[@ratnakantahanse2661]

Thank you sir...

[@ujjwalpandey8212]

Sir can't we use 3 states for this ?

[@vinaypentam1511]

sir i think this is wrong because if the first symbol is 1 then we have to push it but there is no transition to do such operation

[@ammanbaheti3789]

you have not covered the case of string which start with 1 and still satisfies the giiven ccondition

[@lalitkumar8445]

Sir, jb last m € epsilon aaega as input and z0 will be in stack.....then PDA goes to final state.....now input queue and stack will be empty at all or still epsilon and z0 will be exist in both? As no operation will perform for that i guess

[@rustin2380]

Never thought raka zone gaming would start teaching online XDD

[@asimapal8847]

but why is there not more than one state ,like it should move to the next state na.Can you explain it please?

[@codingwale8]

Sir I have a question that if we perform push for zero and pop for one and a string is 0110 then what will be the answer We can push for zero but how we can pop 2 times for 1

[@s2_ultimatetech655]

I think Sir this machine will not work with string '011100' 🤔

[@monikameena6909]

Sir plz provide compiler design classes

[@sahilmishra1999]

Didi phele pda toh pdh loo

[@Developers657]

God mode activated

[@avranj]

why there is no change in state ......we have changed our inputs from 0 to 1 and vice versa ??????

[@aparna8027]

Couldn't understand that either

[@thasleemmd4549]

Sir kyu aapne state change nahi kare input symbol 0 to 1 change hua jab?? Please reply sir

[@sawzeetmaharjan9559]

Y'all, why there is no change of state while doing 1,0->E . He is doing all in one q0 state. kinda confused

[@Parm1.9M]

Is there same process for n(a)!=n(b). ??

[@lakshyadalal8296]

9:16 next transition ka fanda clear hi nhi ho rha Kab jana h next transitions pe!

[@mantavyajain9905]

can we denote (1,0/E) once like we did for (0,z0/0z0) ?

[@shortsspecial842]

1,0/e in loop why you wrote it again 10:13

[@afridnawaz227]

Yeah he did a mistake

[@priyankagandotra5657]

Sb shi hae pr mughe abhi bhi nhi Smj Aya k states change kb krni hae or kb self loop leni hae....🤦‍♀️

[@satyampandey1962]

Same problem

[@MaitohTutgya]

it's incomplete, more transitions are to be included

[@mathflixbyumer2425]

you wrote (1,0 -> Epsilon) twice , was it a mistake ?

[@32subhashreemitra96]

Sir, I'm not understanding when to change the state and when not to?

[@thepriestofvaranasi]

Same. Sir changed states last time but this time he made all the transitions in the same state. I don't get it.

[@shagun3622]

But what if string starts from 1 ...

[@gangadharasarmadantu7363]

Sir, can we design a PDA that accept any no. of strings consists of equal no.of 0,1

[@nehalpatil7]

@GateSmashers what if the string starts with 1

[@harshadadharne-1211]

I understand the solving process but can't get that why there is just 1 state i.e q0 have all operation

[@karanbagle6351]

What if string starts with 1?

[@rifatara8177]

I don't understand when to change state and when to loop! Can anyone tell me?

[@SamridhaRajbhandari]

Is this logic also applicable for 011001?

[@manjotkaur5463]

Sir if string is 1100 then there is initial symbol z0 in stack then how we can perform 1,0/E transition.... Please clear my doubt.... How string will be accepted

[@kamalchhimwal9362]

it will not work for w = {00111100}, threre will be three states

[@UCSDebajyotiDas]

How this machine accept the 10 string

[@Dheeraj-ru6jy]

भैया जी आप से request🙏😊 है की आप datastructure and algorithm ki नयी सीरीज चालू कर दो plzz भैया जी

[@sahilmishra1999]

biiya phele pda toh pdh lo

[@md.borhanuddinhimel7241]

how can we push 1001 in stack?

[@apoorvapathak5888]

why did we made two same transitions??

[@vanshikasharma59]

Didn't undrstnd when the state will chng

[@gunjanarya2915]

Please make more vedios for dsssb tgt... Please🙏

[@omkarsalunke8172]

What if the string is like 011100? It will firstly push 0, then pop 0, then what??

[@ritabratadas8183]

Sir but the solution is not accepting strings like 01100011

[@vinitakeer515]

Sir next to ka video kb ayega???

[@rvr61]

aa jayega!