Conversion of NFA to DFA (Example 2)

[@tanvirhasanmonir1627]

Millions of thanks to you dear sir. Your tutorial is the best material to learn "Theory of Computation"

[@mohdamaan5551]

after 6 years ,still helping passing semester exams.

[@ompadhi1476]

fr brother

[@soumyaranjansamalkiit]

yeh!

[@zarefgamz2515]

very true

[@Ogg_skimmer]

True bro

[@deveshchahar4303]

Passing!! bhai isse padke to top hai

[@capitalofcambrom]

These examples are fantastic. Thanks for explaining it step-by-step!

[@Netugi]

Right? It really helps having every detail explained clearly, so I can see what the process actually is. Was about pulling my hair out doing my class assignment because my notes weren't the best.

[@justinphilip3502]

2 years back this day,best tutorial ever

[@siddhantgarodia3381]

still the best lol

[@assaqwwq]

i love it when he asks "what is the meaning of this?". my gut reaction is of "i dont know, its not my fault"

[@bestof1506]

Is this a reference to something?

[@phoenixultimate8253]

no its just a funny reaction

[@aadisupersonic]

lol.. relatable enough..

[@HARIHaran-ks7wp]

lmao so true

[@AkbarAli-wx5ld]

😂😂😂

[@prasathmsd0760]

All the best for tomorrow's exam mate😂

[@GauravRao09]

Thanks bro my exam is tommorow

[@futurehofer1564]

thanks lol

[@4bdisa]

🤣👍

[@mhmd1759]

lol my exam after 10 hours

[@blacklion9901]

I'm going to exam hall while watching 😂

[@Rahil_Ali_Khan]

ans : should not start with 'ba' and must have odd no of b(s) at the end

[@sgr2683]

correct!! Thanks

[@dheepakraaj8352]

But abab is accepted ?

[@Nishaang]

@@dheepakraaj8352 strings with not starting with ba and having odd no of b's and ending with odd no. of b's

[@paramsavalia7900]

for ababab, it is wrong according to your condition

[@Rahil_Ali_Khan]

@@paramsavalia7900 yeah you're right

[@sahil.sjiwane]

This question came to my university paper. Thanks for the help 🌟

[@siyasahoo4965]

The Q are really helping clearing the concept. Even after 7 years it's still helpful ❤

[@chocolateboyanubhab7340]

i am really a fan of your teaching...such complicated topics are being taught in an easy and simply wap...HATS OFF to you sir...Lots of love from the bottom of my heart.....

[@rajasarodar3628]

NFA : set of all strings over {a,b} that ending with b DFA:set of string b or atleast one 'a' followed by a 'b'

[@maxba2543]

aabb?

[@supersakib62]

Assignment Answer could be: L = {Set of all strings over (a,b) that ends with 'b' but does not start with 'ba' and has odd number of 'b' incase of no 'a' and does not contain 'ababa'} For both DFA abd NFA : {set of all strings over (0, 1) that does not start with 'ba' and ends with 'b' and has odd number of 'b'} 1) ends with an odd number of b’s 2) cannot start with “ba” 3) cannot contain “baba” after an even number of b’s or any number of a’s

[@dakshdixit7286]

bbb is answer according to your logic but it will fail So it think answer should be {b, does not start with 'ba' and ends with 'ab' and has odd number of 'b'}

[@avinashmishra9163]

@@dakshdixit7286 So, what is right task done in this question?

[@dakshdixit7286]

@@avinashmishra9163 i wrote in my comment

[@egetahayamaner7094]

@@dakshdixit7286 How does bbb fail it is in the accepting state.

[@msbrdmr]

I think its {dont start with ba, end with b or ab}

[@akshatgoyal4035]

Great video! Assignment question answer: It takes a string ending with odd number of b to reach it's final state.

[@lokeshyanamandra7775]

But in some cases b Is even bro,so I think so the result will end with b.

[@younisthetrainer8461]

@@lokeshyanamandra7775 no it takes only odd number of b's

[@rohanghauri1935]

try "abbabab"

[@AcezeroGame]

I think it accept every string ending with b

[@robintheunleashed2025]

@@AcezeroGame yeah its string ending with b

[@tomzhang3597]

1) ends with an odd number of b’s 2) cannot start with “ba” 3) cannot contain “baba” after an even number of b’s or any number of a’s

[@mohak23898]

Correct

[@syedqamberalikazmi1001]

You're so genious

[@OllieGee333]

#3 is not correct, "bbabab" is accepted.

[@poonamshekhawat1576]

i.e strings ends with 'a((b)^n+1) ; n={0,1,....}

[@OllieGee333]

Mayurdeep Pathak thats 1 b... thats an odd number of b's

[@SumitKumar-ne9kc]

So answer of the last question is : string must contain odd number of b's after the first 'a' from the right side and also it must not contain odd number of b's before first 'a' from the left side. all string which follow the above constraint will accept by the given NFA please correct me, if I'm wrong.

[@lone_wolf2020]

i think -> The given NFA and it's equivalent DFA accepts any strings made with {a, b} and that must have a single 'b' at the end. try input abab --where no odd numbers of b is present. Feel free to correct me if you think i'm making any mistake.

[@sravanip4509]

@@lone_wolf2020 But ' bb ' is not being accepted..

[@tahaawan7958]

@@sravanip4509 it accepts any string that ends with a non consecutive 'b' at the end.

[@viveksharma5672]

Thanks Neso Academy you are doing a great job for students like us. 👍

[@stefangjorgjevski1990]

answer: case1 : strings can contain only one character 'b' , case 2: start with 'a' followed by as much 'a' as you want and end with an odd number of 'b's

[@fasterbaiter]

it can also accept strings with no 'a's and only odd no. of 'b'. like if you see, it can accept 'b' or 'bbb' or 'bbbbb' and so on..

[@rbek]

@@fasterbaiter yes That is exactly what i was about to say

[@stefangjorgjevski1990]

@@fasterbaiter so it would be 3 cases?

[@fasterbaiter]

@@stefangjorgjevski1990 no still 2 cases. As a single b is included in the case of odd no Of bs

[@fasterbaiter]

@@stefangjorgjevski1990 I think the best answer would be that this FSM accepts string of the form (a^n)(b^m) where n is a whole no. and m is a natural number

[@mirzakariaislam610]

Thanks a lot. Today is my exam. I took my preparation through your lecture.

[@yuvethiekasri606]

Assignment Answer could be: L = {Set of all strings over (a,b) that ends with 'b' but does not start with 'ba' and has odd number of 'b' incase of no 'a' and does not contain 'ababa'}

[@lone_wolf2020]

The given NFA and it's equivalent DFA accepts any strings made with {a, b} and that must have a single 'b' at the end. Thanks for the lecture sir. ---------

[@Akash12-b7n]

incorrect try baab

[@youucanexamprep]

Strings which are accepted in NFA and DFA both are - 1. Only b. 2. End with b. 3. End with ab.

[@anubhavsingh9852]

wrong,, fails for ababab ,bb,baab etc etc

[@rhaypapenfuss]

Neso academy is such a life saver

[@StanleyChidera-c2h]

L = { set of all strings over (a,b) that contain just 'b' or that starts with 'a' and ends with 'a*b' or that starts with 'bb' and ends with 'a*b'}

[@RohitRaj-sf6th]

For both DFA abd NFA, {set of all strings over (a, b) that ends with odd number of 'b'.}

[@karanvenkateshupamanyuise-9407]

it accepts "abab" though. (you've probably already passed atc lol but just saying)

[@stefanif2002]

doesn't accept babb

[@thyagarajarathnampillai5391]

Assignment answer. String cannot start with 'ba', ends with odd number of b's or at most 2 consecutive 'ab's

[@ruili7941]

{a*b, any number of 'a' end with odd number of 'b'}

[@anetaivanicova1709]

it accepts strings that start with a*n, n=0,1,2.. and followed by b*n/2 != 0

[@ProfessionalAmateurQ]

but it can also accept strings that start with 'b', no?

[@amalalzaben3709]

thank you so much you really saved me ,i have an exam next week and i'm ready because of you

[@oceanagencymcp6673]

Case 1: L = { Set of all strings over (a,b) that start with 'b'} Case 2: L = { Set of all strings over (a,b) ending with atmost one 'b'}

[@elisamutiofficial2245]

Thank you Sir. When I learn with your videos , I understand everything thing.

[@sstxixo3975]

My answer: All words ending in "b", can't start with "ba", can't contain "baba" (all applied at the same time) Example words that work: b, a(a)b, bb(a)b, bbbbb(bb), a(a)baa(a)b, bb(a)baab; Example words that break the automat: ababa, ba, a(a)baba; Note: Brackets symbolize periods such as in some representations of rational numbers.

[@nas.a3512]

"bb" is also excluded. Both the DFA and NFA representation do not accept any example words that end with the input "abb"

[@pulkitdhirana215]

Best tutor ever on YT

[@IbrahimAli-mj8ye]

The machine accepts the string 'b' or a string of at least one 'a' followed by a 'b'.

[@dereck2199]

Ans :- 1. it will take odd numbers of 'b' 2. it will take odd number of 'b' end will be followed by 'ab' example :- ''bbb', 'bbbab' this will be accpected by the DFA

[@arsh5587]

Nah bro, abab is also accepted in that dfa , according to me it's the collection of string ends with b

[@dereck2199]

@@arsh5587 i mentioned end with ab.... That abab ends with ab..... So it is accepted

[@arsh5587]

@@dereck2199 yeH u r right but b can be even also

[@jenilpatel5610]

Using Arden's Theorem on Above DFA, I get my regular expression solution as: L = ( (b) + ( a + bb )(a + bb + baa + babb)*(bab) ) Which is 100% correct but help me out with reducing it... To claritfy my answer in first bracket there option between either 'b' or whole right side term. in that term, there's three concatinated strings. 1st string has two options 'a' or 'bb' 2nd string has 4 options which is starred (*) which means it can be used 0 or more times. 3rd string is fixed with 'bab'

[@guoweih7339]

The answer should be: (a +(a+bb)b*a + bb)*b In order to do this, you need to first watch lecture 51.

[@itachi-senpaii]

thanks

[@shumailasaifi]

i first time create diagram by myself feeling good. Thank you sir. Allah-humma-Barik to you are your family.

[@it-a-024-ibrahimfarwahaffa2]

I think he is not a Muslim ....

[@gedelasivakrishna]

The above DFA & NFA are accepting the Strings that end with 'b'

[@yasina63]

I found out NFA is stated based on the current input. The DFA has a unique transition from each state to every input. Thanks a million. ❤️🙏 Today I have a final exam. I hope I will get an A+

[@navabshaik7280]

It's a dad of all channels to understand kinly the Automata &TOC

[@gautamkatyal9966]

It accepts the String start from a and nth number of a's and ends with b or String starts with b followed by b and nth number of a's and end with b.

[@rohitnair9022]

Why b and not nth number of b's?

[@MrLarryLicious]

THANK YOU FOR YOUR EXISTENCE

[@rajashekar4171]

Hi

[@pranayaghimire1122]

Neso academy is heart of csit students.

[@RoshanKumar-d8o]

Bhai aap kon se college se the? Or aapka college me placements ka kya scene tha/hai, Or aapka placement hua?

[@asefahmed11]

After 6 years, still helping me

[@BillPark-ey6ih]

This guy saved my life.

[@MischievousSoul]

It's ans is the strings that ends with odd no of B Because it accepts every string that has odd no of B and doesnt focus on the no of a and b it had in between

[@anubhavsingh9852]

string abab is accepted.what now ;-

[@RittickJana]

Thank you so much.... I express my sincere gratitude to you.

[@FzOmlan]

Answer: The NFA accepts strings that contain at least one 'b', or sequences of 'a' followed by 'b' This DFA accepts strings where the input ends with the character 'b'.

[@phantomv1951]

L={set of all string over 'a' and 'b' that ends with 'ab'}

[@SaurabhGupta-sx4jy]

L={Set of all strings over (a,b) that ends with 'b' , 'bbb' or 'ab'}

[@vrushaligholap6647]

GREAT EXPLANATION SIR! YOU MADE IT VERY SIMPLE AND EASY.

[@arunkumarsundaram9941]

Accepts a string ending with odd number of b's unless one of the following is true: 1. String starts with 'ba' 2. There is an occurrence of 'aba' after an odd number of b's

[@alan2211]

No, strings starting with 'ba' will end up in the dead state.

[@TariqueGauri]

This condition is not true for aabbb

[@himanshugupta3028]

It accepts the string which satisfies following conditions: - 1. It should not start with 'ba'. 2. Total Number of b's in the string should be odd. 3. The string should end with odd number of b's.

[@korhandenizakn6107]

Are you sure with your 2. sentence. Try "abaab".

[@niteshmodi5468]

abab is accepted

[@randomstuff5714]

ASSIGNMENT :- a string which starts with even number of 'b' and ends with odd number of 'b'

[@rajearyan]

i dont think it is even number of b's. It is odd only.

[@okko7788]

But abab works but shouldn't be accepted according to your definition

[@randomstuff5714]

@@rajearyan No, It must be even also. Example- "ba" won't be accepted but it still contains odd number of b's.

[@randomstuff5714]

@@okko7788 No, it will be accepted according to my definition. "abab" is starting with 0- 'b' and zero is an even number and it is ending with 1- 'b' and one is an odd number.

[@randomstuff5714]

I can see that my definition is wrong. one condition is also there, that after odd number of 'b' there should not be 'aba'.

[@sagarr3964]

It's 2021 , and tomorrow is my exam I have no worries after watching these videos

[@danteeep]

far the best video explanation with an example for the conversion of NFA to DFA. thank you so much for this !!!!!!

[@RoshanKumar-d8o]

Bhai aap kon se college se the? Or aapka college me placements ka kya scene tha/hai, Or aapka placement hua?

[@aniketkumar9621]

string that accepts all the string over {a,b} that starts from a & must end with ab

[@mozammalhossain968]

The language accepts strings 1. Doesn't start with 'ba' AND 2. Odds number of 'b' in sequence from the ends

[@sahilnaik3079]

That is not correct according to your explanation ababab should be accepted. But your answer is close. Some modification is required

[@raviranjan1810]

your videos I just awesome, no words god-level explanation with full concept

[@abdullah-al-foysal3731]

Assignment solve: Regular expression for accepting state over input symbol (a,b): bba*ba | aa*ba | b Accepting Language grammar G={(S,A,B),(a,b),(S),(S=>bbB | aB , B=>Aba ,A=>aA | ε)}

[@pranavagarwal4827]

the string that only ends with odd number of b is allowed and if the string starts with b and has to contain a , then their must be at least 2 b in the start

[@kingbond470]

On condition and---- ab🤔?? And for odd number abbbab??

[@shivamjaiswal7950]

@@kingbond470 the string that contain odd no. of b at end

[@kingbond470]

@@shivamjaiswal7950 abbbab??

[@utshodas001]

great video. is the answer " a^n b^2n+1 n>=0 "?

[@shanmugapriya7554]

L={set of all strings with 0 or a's and end with odd number of b's}

[@AnimeManiaa]

String that contains any no of a's followed by 0 or even no of b's and end with odd no of b's

[@jay3101]

for NFA sol :accepted string ending with. (b) followed by even no of (a).

[@VishalSharma-ks6ip]

ab is also a string which is accepted.... Here a is not even

[@hesham.abd-allah]

The assignment's solution is: L={wb^k | w belongs to {a,b}* & k%2=1(k is an odd number)} In other words... it accepts all strings that end with an odd numer of 'b's

[@in_sight2236]

incorrect: what about babbb? it ends with d state.

[@shrunkhalawankhede2611]

VERY NICE EXPLANATION. NICE ACADEMY ROCK IS ALL SUBJECTS

[@TEMSGENGETIE]

it is very very very good tutorial live in ethiopia

[@aayushve426]

Answer to the assignment: set of strings on (a,b) such that no 'ba' string is present and odd no. of b's at end of string.

[@frankhilgenberg7310]

ASSIGNMENT: L = {Set of all strings over (a, b) that ends with a odd number of 'b's}

[@anubhavsingh9852]

check it for ababab

[@frankhilgenberg7310]

ups! With ababab I will end in state D and this is not a final state. So - let me think about it again ;-)

[@hariprasath4663]

Assignment: Both the NFA and DFA can accept strings that end with Odd Number of b's, and it will not accept strings, that start or end with ba

[@yahyairfan1159]

Starting with a and ending with odd b,s OR starting and ending with a 'b' only

[@MyAsdfqwe]

this and union, starts with even b's followed by infinite a's with even b's and ends with b.

[@Viggen66]

Thanks for this teaching, awesome lecture

[@yuyao4008]

way much better than my lecturer thanks a lot.

[@adityakukade3691]

all strings ending with b or odd no of b's or a followed by odd no of b's

[@deepakvaishnav9780]

It is accepting the strings that end with sequence of 'b' having odd numbers of b consecutively

[@fasterbaiter]

I think the best answer would be that this FSM accepts string of the form (a^n)(b^m) where n is a whole no. and m is an odd natural number

[@babyharsha9162]

For DFA, a string with only 'b' that are odd in number Start with 'a' ends with 'b' Starts with 'bb' ends with 'b'

[@anubhavsingh9852]

what if string is ababab fails for case 2

[@ZeeshanAliQureshi93]

b(bb)^* + a^+ b(bb)^* String with odd numbers of b OR a string which starts with a(min one 'a') and ends with odd number of b.

[@thrinadhreddyjonnalaAP2211]

Video can be old but knowledge cant ...( 8 Years still helps.)🤘

[@janvieriyakaremye7694]

L is a set of strings over a and b that ends with b

[@shubhrajyotipoddar1684]

set of all strings starts with {a, b} that ends with odd no of b

[@longlivefreedom]

Sometimes I found your videos more useful than university lectures.

[@AnukiranGhosh]

all the time

[@Nomnomnom165]

all the time 😂

[@markxavior]

Excellent tutorial professor.

[@buzzfeedRED]

YOU MUST TELL THE ANSWERS OF THE ASSIGNMENT YOU HAVE GIVEN IN YOUR VIDEOS ,

[@sundarraj6509]

I think it’s answer is that “the string that either end with ‘b’ or ‘ab’”.

[@Adithyan7274]

@@sundarraj6509 ​you are correct but there's one more.. the number of ' b ' should be odd, because even number of ' b ' ends in the state AB which is not the final state.

[@ganeshpandeyKing_Brian]

@@Adithyan7274 The machine accepts string "abab" so i guess there's more to it than just odd numbers of "b".

[@abelii8726]

i think if the number of 'b' is either prime or odd then it's accepted by the machine.

[@yokeshnsr1976]

@@abelii8726 but 'abbbab' is also accepted

[@bellajames7913]

Set of all strings with L =( a^n)b

[@prashanthi8492]

the answer to the last question is the string that contains only odd no.of b's, strings that ends with oddno.of b's.

[@navaneeth2000-qc]

Thankyou very much for this valuable class

[@tolulopemalomo8922]

I think the DFA/NFA accepts all strings over { a , b } that end with an odd number of b-s. Please let me know if I'm right.

[@SahibpreetKaur]

I think it's not odd number of b-s but rather all strings ending with a single "b". For example: b, ab, aab, abab. But NOT abb or aabb or aabbb.

[@amitpatelyt12]

❤❤❤ while(i=1) { printf("Thank You Sir "); }

[@darraghmanning]

The string must satisfy all 4 of these conditions: 1. Not start with BA 2. Not contain BABA unless the preceding substring contains an odd number of B’s 3. Must end in B 4. Must have an odd number of B’s following the last A Can anyone confirm if this is right I spent like 40 mins thinking about this haha

[@thyagarajarathnampillai5391]

Assignment answer. String cannot start with 'ba', ends with odd number of b's or at most 2 consecutive 'ab's

[@abanoubbenyamin1729]

you saved so many students. Really god bless you

[@snehalmali9630]

Strings will be accepted which doesn't start with 'ba' and containing odd no of 'b's and end with either single 'b' or 'a' followed by 'b'. Is it the right answer, Sir?

[@ArunKumar-mv3qj]

I think answer will be, accept any string over {a, b}, that ends with 'ab', except starting string 'ba'. Is it right??

[@ajaykarthik01]

thankyou please complete the compiler design tutorial as soon as possible

[@djmaster1110]

Very well explained

[@saurabhmourya7271]

L: {accepts the strings which neither start or end with 'ba'}

[@AviralBangar777]

very good video easy explanation thank you

[@swatimaurya9405]

Seriously amazing you are 😊😊😊can you please tell which editor you use 😅😅

[@RahulKumar-yc2ip]

This NFA (or equivalent DFA) accepts all strings over {a,b} that end with odd number of b's.

[@deniztopay306]

babbb is not accepted. It stays on state D. But all strings that ends with odd number of 'b', doesn't starts with 'ba' and doesn't contain 2'ba's in a row after started corrects this. I am not sure if mine is correct tho.

[@sandipdas5684]

Your tutorial on this subject is best sir... Respectfully thank you so much sir...

[@healthandenvironment2998]

Amazing Sir!! You deserve a subscriber. I HAVE SUBSCRIBED