11:34 Ek dm se inhone waqt badal diya, zazbaat badal diye, zindgi badal di.... Ohhh bhai.... Literally 2 minutes ke liye me stunned ho gya tha.... Itna dimaag koi kese lga leta hai.... Sir Respect++;
I think this may help if youre confused about the last three For 21 and 22, if w and w^r are any other values instead of null, the machine fails. But for say wcw^r, if w and w^r are null, then c could be any word, and the entire combination will basically be c. So wcw^r becomes set of all words(c can be any word). We do have finite automata which accepts all words. What if w and w^r werent null? Then imagine w=abc, c=d, then wcw^r will be abcdcba. It is equivalent to w being null and c being abcdcba. So it can be converted to an equivalent form where w is null and c will be the entire string then, which is acceptable by fsm.
@Tech RK I agree. I have no idea why he would arbitrarily decide to make w null. How does he get to decide that? What if w isn't null? I just cannot understand this.
sanchit sir , I can't afford your course on unacademy plus so please put some videos on various gate cse subjects so that student like me from poor background can get knowledge because your teaching is as good as that i can't explain.......lot of respect sir .
thanks a load sir your videos are really very helpful waiting for the complete series...sir the way you teach makes any subject very interesting and easier ...
Yes Sahil, we are trying to upload videos as frequently as we can. Stay connected !! Keep supporting !! Do visit our website www.knowledgegate.in for more courses and contents.
thank u sir for explaining all types of question that belongs to exam ...............thats make easier to understand topic and also then help for solving exam previous year question ............... .i really want please u upload also algorithm subject becz ur explaination is very helpful for us
In question 21 & 22 : If we consider w and w^r as null. Then the whole string will be null. And we can make finite automata for null string. Then it can also be regular.
For 21 and 22, if w and w^r are any other values instead of null, the machine fails. But for say wcw^r, if w and w^r are null, then c could be any word, and the entire combination will basically be c. So wcw^r becomes set of all words(c can be any word). We do have finite automata which accepts all words. What if w and w^r werent null? Then imagine w=abc, c=d, then wcw^r will be abcdcba. It is equivalent to w being null and c being abcdcba. So it can be converted to an equivalent form which is acceptable by fsm.
Sir in Q23, some unacceptable strings are also accepted (according to the logic you gave). But this reduces consistency. The strings that are not to be accepted are also being accepted. Is this okay/ allowed ?
In the question 23. If we assume "w" is of SINGLE-LENGTH , therefore it's reverse " w^r" would be the same , hence no such AUXILIARY MEMORY is needed or required in that case case also the LANGUAGE be also considered as " REGULAR LANGUAGE".
@@isheep9025 Still unclear. How do you clearly identify ω and ω(r). There will still be cases where ω is not null. I get that c is easily dealt with. Do you have any resource or explanation for it?
1st WCWr; wE Sigma* Here if we consider W and Wr as Empty String, then can we not have c left in the Regular Expression, as a result of which we would just be having one synbol to be accepted, which may be a or b, Can we not make an FA for the same?
Sir you said that this machine accepts every string and we are also considering 'c' as the complete string, but then it will also going to accept other strings like "aabb" which is not of the pattern 'wcwr'. How exactly this is working?
given any string, you can take that whole string as c and assume w and wr as null, and then make the automata for c, thus proving that the language is reg. You have to take w and wr as null, just for the sake of proving that the language is regular
Arjun jha dear arjun brother .. Yhi baat mujhe bhi puchhni thi... But apne puchha aur virat sir ne Jo reply kiya usse mera doubt clear ho gya... Ap dono ko thank you
Sir I have a doubt . JB wcw^r me jo DFA bnaya hai Apne wo un string Ko reject bhi krni chahiye jo wcwr pattern Ko follow nhi office. But jo DFA aapne banaya hai wo to sare hi string accept kr rhi hai.. to kya wo DFA banana Shi hai?
Sir plz response me in all the questions suppose consider Q17 why can't we make it's finite automata? just consider w as 'aaa' then this problem is quite similar as 'aaaSaaa' that means we have to make a finite automata which accepts the strings starting with 'aaa' and ending with 'aaa', so we can easily make it's finite automata. Again similarly in Q20 we have w which can't be done null but what does it need to do it null we can simply suppose w=aab then the string is 'aabSaab', which also quite similar to the problem of making the FA which accepts the string 'aabSaab'
Thanks a lot dear😊 ❤ Keep learning & supporting! Stay connected for more amazing videos & contents 👍 You can also check out our CS bundle course for complete preparation of Computer Science subjects, sharing course link below: www.knowledgegate.in/learn/Zero-to-Hero-in-Computer-Science
Sanchit Sir, In Q23, Q24, Q25, what if w and wr is not null ? Then the generated language is not regular....So language will not be regular in all the possible cases ? Right ?
sir is logic k hisab se 18 numbr m bhi w and wr ko null maan lo, string separator c symbol only k liye machine bna skte h,,,, null ek m hi kyo sbme bhi maan lo, same logic se sb regular languages prove ho skti h,,, plz help and clear this doubt sir
In case you take this in Q 17 then remaining string (other than epsilon ) can not be acceptable , The clue is that when you assume "w" as null then still there should be way left to satisfy every other string of kleen closure which is justified with c as a member of the kleen closure as in Q 23,24,25 , So in every other Q (17 to 22) when we like to take w as null we do not have scope to represent other strings of Kleen closure as a criteria of Regular language.
sir agar {w.w^r | w- sigma *} , isme hum w ko null lele toh bhi toh ek string accept hogi , agar poistive closure hoga toh nhai hogi but klene closure mein toh hogi , sirf ek hi hi hogi par hogi . then woh regular maane ya nahi. ?
in q 23 we are only talking about the case when w is null but according to conditions given w belongs to kleene star what if w is not null? then machine cant accept the language then how can we say it is a regular language????? but only a particular of it is satisfying the need of a regular language
In 23,24,25 questions instead of kleen clouser it will have positive clouser then that language is regular because positive clouser can't accept episoln, plz reply to this dought sir......
nikunj mochi so if c does not belong to universal set than it means not even pda can solve it... if pda can than how is it possible since c does not belong to universal set... can we draw nfa that ignores any other symbol which does not belong to universal set like c
sir you have make a m/c the can accept wcwR| c,w belongs to sigma *, but sir m/c which you designed can accept any string does mean it is palindrome or not, but we are creating palindrome too using this language . So how can you say that this m/c you designed is correct, it is incorrect .
sir in Ques 23 you have taken w to be epsilon and prove that its regular.But if it is regular it should hold for all w belongs to kleen closure.Sir can you prove that it is regular for w not equal to epsilon
In case you take this in Q 17 then remaining string (other than epsilon ) can not be acceptable , The clue is that when you assume "w" as null then still there should be way left to satisfy every other string of kleen closure which is justified with c as a member of the kleen closure as in Q 23,24,25 , So in every other Q (17 to 22) when we like to take w as null we do not have scope to represent other strings of Kleen closure as a criteria of Regular language.
For complete Theory of Computation, check out this video:
th-cam.com/video/9kuynHcM3UA/w-d-xo.html
11:34 Ek dm se inhone waqt badal diya, zazbaat badal diye, zindgi badal di.... Ohhh bhai....
Literally 2 minutes ke liye me stunned ho gya tha.... Itna dimaag koi kese lga leta hai.... Sir Respect++;
exactly bro
Bhaiya ye batiye kya ye rule hum question no 17 pe apply nahi kar sakte h ?? Is video ko check kar batiye.
I think this may help if youre confused about the last three
For 21 and 22, if w and w^r are any other values instead of null, the machine fails.
But for say wcw^r, if w and w^r are null, then c could be any word, and the entire combination will basically be c. So wcw^r becomes set of all words(c can be any word). We do have finite automata which accepts all words. What if w and w^r werent null? Then imagine w=abc, c=d, then wcw^r will be abcdcba. It is equivalent to w being null and c being abcdcba. So it can be converted to an equivalent form where w is null and c will be the entire string then, which is acceptable by fsm.
11:17 onward 180 degree twist
Salute sir...@11:34 i am writing like it's non-RL and suddenly u make twist in answer....your teaching power is great.
@Tech RK I agree. I have no idea why he would arbitrarily decide to make w null. How does he get to decide that? What if w isn't null? I just cannot understand this.
sanchit sir , I can't afford your course on unacademy plus so please put some videos on various gate cse subjects so that student like me from poor background can get knowledge because your teaching is as good as that i can't explain.......lot of respect sir .
Are u from WEST-BENGAL ?
thanks a load sir your videos are really very helpful waiting for the complete series...sir the way you teach makes any subject very interesting and easier ...
most useful video on regular languages on youtube
Sir i am so so Thankful to you, You are a lifesaver. Thank you so much for such an amazing lecture and explanation on this topic.
thanks sir keep going for complete TOC these month.
yes happy, will try my best...
I love you so much sir..
thanks for these videos , one the best videos for gate CSE , may God bless you and your entire family :)
Sir please upload videos on regular basis, it is beneficial for us, We are with you to like your videos.
Yes Sahil, we are trying to upload videos as frequently as we can. Stay connected !! Keep supporting !! Do visit our website www.knowledgegate.in for more courses and contents.
Superb ❤ explanation sir ji.😊
Thank you, Pavan. Keep learning, supporting & stay connected for more content👍
thanks a lot sir your videos are really very helpful
thank u sir for explaining all types of question that belongs to exam ...............thats make easier to understand topic and also then help for solving exam previous year question ...............
.i really want please u upload also algorithm subject becz ur explaination is very helpful for us
In question 21 & 22 :
If we consider w and w^r as null.
Then the whole string will be null.
And we can make finite automata for null string. Then it can also be regular.
My doubt is also same. If you got any solution please explain.
For 21 and 22, if w and w^r are any other values instead of null, the machine fails.
But for say wcw^r, if w and w^r are null, then c could be any word, and the entire combination will basically be c. So wcw^r becomes set of all words(c can be any word). We do have finite automata which accepts all words. What if w and w^r werent null? Then imagine w=abc, c=d, then wcw^r will be abcdcba. It is equivalent to w being null and c being abcdcba. So it can be converted to an equivalent form which is acceptable by fsm.
15:09 yes sir wada kiya, yaad rakhenge 😁😁
Sanchit sir please upload lectures of Compiler Design, SR, LALR parser.. no one can explain all with such an enthusiastic way .. thanks a lot
Sir in Q23, some unacceptable strings are also accepted (according to the logic you gave). But this reduces consistency. The strings that are not to be accepted are also being accepted. Is this okay/ allowed ?
Thank you...I was searching for these example everywhere...thank you so much
You're welcome, @jyotipritamsingh. Keep learning, supporting & stay connected for more content👍
sir please please upload PDA and turing machine at the earliest!! please sir!
and thank you so much for this video. u teach best
working on it shweta...
M new to this channel m glad to see this channel thank you so much sir.... It's helps lot to xam......
vaada raha sir 👍👍 15:11
In the question 23. If we assume "w" is of SINGLE-LENGTH , therefore it's reverse " w^r" would be the same , hence no such AUXILIARY MEMORY is needed or required in that case case also the LANGUAGE be also considered as " REGULAR LANGUAGE".
Sir, considering 23 as Regular. Can we make w, w^r as Null in case no: 17 and w, w as NULL in case number: 18 to make them regular?
c doest belong to sigma star , so u cant expand c here
@@vinayak186f3 pehle apne concept sai krle
@@isheep9025 Still unclear. How do you clearly identify ω and ω(r). There will still be cases where ω is not null. I get that c is easily dealt with. Do you have any resource or explanation for it?
thanks for keeping updated
shukriya ajaz bhai
dhanyavad sir ji😆
best class for TOC
Thanks a lot @debottamkar2364. Keep supporting & stay connected for more content👍
1st WCWr; wE Sigma*
Here if we consider W and Wr as Empty String, then can we not have c left in the Regular Expression, as a result of which we would just be having one synbol to be accepted, which may be a or b, Can we not make an FA for the same?
Thank you very much. You are a genius.
You're welcome dear 😍 Keep learning & supporting ! Do visit our website www.knowledgegate.in for more amazing courses & contents 👍😊
Sir you said that this machine accepts every string and we are also considering 'c' as the complete string, but then it will also going to accept other strings like "aabb" which is not of the pattern 'wcwr'. How exactly this is working?
In question 23 is it possible that it's Automata can accept strings which doesn't belongs to that aceeptable criteria of L ? Please Help !!
Sir, if we change the state of c to the initial. then we can make 2nd a regular language
Sir in 23, if we come across a string abb(a+b)^*abb then this should not be recognized by the FM but here it is been recognized by (a+b)^*
Excellent!!
Love you Sir !!👍
question 23 is 👌
Dear sir you are considering specific as w and wr as null,
otherwise it is not regular
how ??
given any string, you can take that whole string as c and assume w and wr as null, and then make the automata for c, thus proving that the language is reg. You have to take w and wr as null, just for the sake of proving that the language is regular
Arjun jha dear arjun brother .. Yhi baat mujhe bhi puchhni thi... But apne puchha aur virat sir ne Jo reply kiya usse mera doubt clear ho gya... Ap dono ko thank you
but how those languages which has w not equal to null are regular..how machines does remember string w in that case ?????
@@viratmishragta4 bro cant the same logic be applied to q21 , w belongs to kleene closure so w can be null there also and null is regular..
@@alihashmi9519 if you take w as null, then ww will be null, then how will you represent the strings? (Like abab, where w=ab)
Sir if in question 23,24 and 25, it is positive closure instead of kleen closure, whether language is regular or not?
Sir I have a doubt . JB wcw^r me jo DFA bnaya hai Apne wo un string Ko reject bhi krni chahiye jo wcwr pattern Ko follow nhi office. But jo DFA aapne banaya hai wo to sare hi string accept kr rhi hai.. to kya wo DFA banana Shi hai?
15:15 ye dono languages bhi kya hain? *regular*
Not *non-regular*
I dont understand how these languages are regular how would we match w and wR . If we do what sir told then what us the point of saying w and wr?
Sir plz response me in all the questions suppose consider Q17 why can't we make it's finite automata? just consider w as 'aaa' then this problem is quite similar as 'aaaSaaa' that means we have to make a finite automata which accepts the strings starting with 'aaa' and ending with 'aaa', so we can easily make it's finite automata. Again similarly in Q20 we have w which can't be done null but what does it need to do it null we can simply suppose w=aab then the string is 'aabSaab', which also quite similar to the problem of making the FA which accepts the string 'aabSaab'
Hats off to u sir
Thanks a lot dear😊 ❤ Keep learning & supporting! Stay connected for more amazing videos & contents 👍 You can also check out our CS bundle course for complete preparation of Computer Science subjects, sharing course link below:
www.knowledgegate.in/learn/Zero-to-Hero-in-Computer-Science
Thank's Sir......
these last three questions were really hard to guess these type of out of the box thought process is really hard to crack no matter how much i study
sir in q21 , w can be null as it belongs to KLEENE closure so ww = NULL and null is regular so inst q21 regular lang?
Sanchit Sir, In Q23, Q24, Q25, what if w and wr is not null ? Then the generated language is not regular....So language will not be regular in all the possible cases ? Right ?
my ques is same have you got answere ?
sir is logic k hisab se 18 numbr m bhi w and wr ko null maan lo, string separator c symbol only k liye machine bna skte h,,,, null ek m hi kyo sbme bhi maan lo, same logic se sb regular languages prove ho skti h,,, plz help and clear this doubt sir
Sir if w is sigma* then wcw^r if we make w =€ then w^r is € then c is just a symbol then wcr can be accepted can u plz tell me
Sir plz make video on how to find the language is cfl,dcfl,csl....
Self referential note:
Q-23,24 are trick questions
sir if W is taken null in problem no. 17 then it will regular as you said in case of 23,24 and 25
In case you take this in Q 17 then remaining string (other than epsilon ) can not be acceptable , The clue is that when you assume "w" as null then still there should be way left to satisfy every other string of kleen closure which is justified with c as a member of the kleen closure as in Q 23,24,25 , So in every other Q (17 to 22) when we like to take w as null we do not have scope to represent other strings of Kleen closure as a criteria of Regular language.
sir.. little confusion on 23, 24 and 25
a finite machine should accept all possible strings so why we assuming that w will be null only...
sir how are you making w and wr null!!
u will still need to know the start of wr and end of w atleast!!
please help :)
ping me at mail.knowledgegate@gmail.com
NIKUNJ MOCHI Nice one that made me understand the logic Thank you so much
sir agar {w.w^r | w- sigma *} , isme hum w ko null lele toh bhi toh ek string accept hogi , agar poistive closure hoga toh nhai hogi but klene closure mein toh hogi , sirf ek hi hi hogi par hogi . then woh regular maane ya nahi. ?
Thanks Sir but also uploaded some videos for other sub as computer network, AI not only gate Point of view but UGC net also
in 1 case can we consider w as null and c as any string it should also be regular then
sir please try to clarify more practically.....and by some methods like making it's machine or by pumping lemma
in q 23 we are only talking about the case when w is null but according to conditions given w belongs to kleene star what if w is not null? then machine cant accept the language then how can we say it is a regular language????? but only a particular of it is satisfying the need of a regular language
Sir Hmari teacher ki mataji ki 😺😺 :)
we can consider w and wr as null in other examples also then why they are nonregular...
please sir upload more videos of TOC
In 23,24,25 questions instead of kleen clouser it will have positive clouser then that language is regular because positive clouser can't accept episoln, plz reply to this dought sir......
Ak question hai 23 ,24,25 main c,w dono hi sigma* say belong kr rhi hai aghr w 1 yah kuch bi lay phr bhi is ka finite automata bn jay gaa ?
Why we can not assume w as null in firsr cases
Ur videos on cfl and csl
Sir i am learning from your video for my university preparation
Can you recommend a good book that can cover theory well
Hi dear, you can refer KLP Mishra's book.
@@KNOWLEDGEGATE_kg ok thankyou sir
Sir please explain pumping lemma!!
Sir...
Jis tarah 23 me maan liya ki 'w' ko hm null le lenge. Usi tarah 1st question me v to hum 'w' ko null le to wo v to regular ho skta h?
Ok..
But sir, can't we transit from initial state to final state with 'c' symbol?
Ok
Got it.
Thanks sir...
nikunj mochi
so if c does not belong to universal set than it means not even pda can solve it...
if pda can than how is it possible since c does not belong to universal set...
can we draw nfa that ignores any other symbol which does not belong to universal set like c
excuse me sir, in Q-17 & 18 kya hum w ko null nahi bana sakte ?
In question 17 if w=epsilon then i will change state after seeing c and accept every thing since w and wr is epsilon
Sir, what if we do not take w = null in last 3 questns
Thankyou sir
Can't we use pumping kemna here
What will be the case if L={wCw^r | C,w belongs to x^+} Where x is sigma+.?
Without keeping C in universal closure, why can't we still make w and wr empty and make language equivalent to L = {C}
Sir agar hmm eq 17 me bhi w and w{reverse} ko null treat kre tb to 17 eq bhi to regular language prove hoo jayega
in question 17 C is not belong to sigma*
is english version available?
you again spoke "non - regular" in the end instead of " regular ".
sir but in toc string not be null in question 23 how can you take w is null
sir you have make a m/c the can accept wcwR| c,w belongs to sigma *, but sir m/c which you designed can accept any string does mean it is palindrome or not, but we are creating palindrome too using this language . So how can you say that this m/c you designed is correct, it is incorrect .
sir in Ques 23 you have taken w to be epsilon and prove that its regular.But if it is regular it should hold for all w belongs to kleen closure.Sir can you prove that it is regular for w not equal to epsilon
I believe something is wrong with ur play list as it's not showing all videos
Sir I want to know if your course is enough for University exams ??
Yes
Sir if we assume Q17 w is Null because w€universal set
In case you take this in Q 17 then remaining string (other than epsilon ) can not be acceptable , The clue is that when you assume "w" as null then still there should be way left to satisfy every other string of kleen closure which is justified with c as a member of the kleen closure as in Q 23,24,25 , So in every other Q (17 to 22) when we like to take w as null we do not have scope to represent other strings of Kleen closure as a criteria of Regular language.
L={waw^r | w belongs to kleen closure and (a,b) are alphabets}
toh ye language bhi regular hai as per 23rd question.
m r right sir?
Thank You Sir
Most welcome dear! Keep learning & stay connected! You can also check out our website www.knowledgegate.in for more amazing videos & contents👍😊
WCW= World Championship Wrestling!!!
at 6.54, duster was bit sad. :-P
#imp 23 onwards ques
Your videos are amazing sir but can you explain in English? That will be more comfortable
Sir please upload videos on DAA, DATA STRUCTURE, ADBMS and Unix linux Subjects in Hindi sir.... Please Upload Hurry UP Sirrrrr
Jo trick que no. 23 or 24 m apply hua wo que. No. 17 m kyu nhi hua
24 , 25 regular hai apne tick to kiya butbbola non-regular
L= { wcw^r | w=0 }
would this be regular lang or not?
yes if length of w is 0
KNOWLEDGE GATE thank you!
@@KNOWLEDGEGATE_kg then sir Q17 i.e., L={wcw^r | w € all} should also be regular
Sir please aap hamare comments bhi check Kar liya Karo Kabhi
It's our humble request 🙏🙏🙏🙏🙏🙏🙏🙏🙏
TH-cam needs a 3x
Bhai yahape paid course se achha padhaya hai 😢
Yo
Waise to 18 wala bhi regular hai w ko null Maan lenge
@@akshaydagar9911 Stupid just because he asked a question which has been explained in a lousy manner in the video?