inception movie is over rated in front of arden's theorem's proof
@sohamshinde1258
3 жыл бұрын
@Asher Blaine U are lying, the website which you mentioned is a scam which demands payment for the hacked account's login details and once we enter about payment details there, they hack our bank account and loot us and we never get any hacked info of any account.
@chandhunukala6949
3 жыл бұрын
@@sohamshinde1258share they website link to me
@sohamshinde1258
3 жыл бұрын
@@chandhunukala6949 He removed that commment I didnt had it saved :(
@nithinkumar6105
3 жыл бұрын
🤣🤣🤣
@tusharupadhyay1356
4 жыл бұрын
I used the stones to destroy the stones😂
@griotolu7040
3 жыл бұрын
lmaooo
@goalshola5628
2 жыл бұрын
Yees he proved it with what he has to prove. Worst
@toomanyarguments
2 жыл бұрын
mann that is really true 😂
@NINJaGAMINg-dr6uf
9 ай бұрын
😂😂 marvel fan everywhere... Thanos was right 😂
@riturajnavindgikar1529
2 жыл бұрын
Correct is R = Q + RP … (1) R = Q + (Q+RP)P …(2) from 1 R = Q + QP + RP^2 Again keep substituting n times R = Q + QP + QP^2 + QP^3…. R = Q( E + P + P^2 + P^3..) R = QP* Proved
@sourabhshekhawat2364
Ай бұрын
thankss broo ✨👑
@pritam_shejul
6 жыл бұрын
2 min of silence ....what a proof
@abhishekkumarsingh9938
5 жыл бұрын
same here bruhhhhhhhhh
@cyanfroste5559
4 жыл бұрын
`I used the R = QP* to derive the R = QP*`
@ru2979
4 жыл бұрын
Idiotic proof
@hmm7458
4 жыл бұрын
@@cyanfroste5559 oo i see a man of culture
@marxman1010
3 жыл бұрын
@@cyanfroste5559 Derive the uniqueness, but why it means unique answer?
@zackcarl7861
2 жыл бұрын
Normally in regular algebra or , trigonometry n, when in an equal its given an equation- x=y+x+1 And if we say we put x= -y we put LHS x, and RHS x both as -y and get answer like y=1 but here in ardens theorem we , did not substitute value for LHS ,R
@sknasimhossen9546
9 ай бұрын
Mindblowing Explain ability of Nesco academy.Thank you sir. ❤
@NINJaGAMINg-dr6uf
9 ай бұрын
😂
@SumitKumar-fj9sy
3 жыл бұрын
Sometimes sir's Genius… It's Almost Frightening.
@espio3364
2 жыл бұрын
XD It's funny what he did but that one thing makes me remember Arden's Theorm forever
@an_archy
Жыл бұрын
his genius has a gravity of its own
@abhayrajlodhi4949
27 күн бұрын
How to fire a fire take out the fire from fire😂😂😂
@abhaylodhi7279
27 күн бұрын
you funny person
@VijaykumarVijayKumar-nq5yl
6 ай бұрын
Love ur teaching bro❤
@learnwithmanu5655
7 жыл бұрын
Thank you sir.and sir please post some problem on Arden theorem and how to convert a regular expression into finite automata.please sir .
@abdulhaseeb2966
3 жыл бұрын
Statement − Let P and Q be two regular expressions. If P does not contain null string, then R = Q + RP has a unique solution that is R = QP* Proof − R = Q + (Q + RP)P [After putting the value R = Q + RP] = Q + QP + RPP When we put the value of R recursively again and again, we get the following equation − R = Q + QP + QP2 + QP3….. R = Q (ε + P + P2 + P3 + …. ) R = QP* [As P* represents (ε + P + P2 + P3 + ….) ] Hence, proved
@dailymemes2512
3 жыл бұрын
And what about R inside the expression where it is gone
@zackcarl7861
2 жыл бұрын
@@dailymemes2512 he put the value of r agin and again
@kaushalkumar1664
2 жыл бұрын
this is right. In the video, that guy is proving using the statement that is to be proved.
@sleepypanda7172
2 жыл бұрын
Much better. I can sleep in peace now
@rfyl
Жыл бұрын
@@kaushalkumar1664 The *intention* is quite clear and correct: the number of P's can increase to any arbitrary number -- including none at all in the first "Q" -- so they are "becoming" P*. But yes, strictly speaking he is using circular reasoning in that very last step. The solution is a very small but important change to the proof: rephrase it as a proof by induction. That way "R = QP*" is introduced (correctly) as the Induction Hypothesis, rather than circularly as a "fact".
@HumphreyTembo-d4p
4 ай бұрын
I KNEW NESO IS THE BEST EVER TUTOR
@lamaspacos
5 ай бұрын
05:50 R = Q + QP + QP^2 + ... + Q^n + {some strings with length > n ----- unless trivial case P empty}, for all n \in N Then R = Q + QP + QP^2 + ... = Q P*
@lamaspacos
5 ай бұрын
05:50 R = Q + QP + QP^2 + ... + Q^n + Kn, where the later is defined as RP^(n+1), for all n \in N. Notice that the intersection of each pair (Kn1,Kn2), with n1, n2 \in N, is empty. Therefore, R = Q + QP + QP^2 + ... = Q P*
@astha_25
5 ай бұрын
Amazing explanation ❤
@nagapushpa1041
4 жыл бұрын
Thank you very much sir Good explanation
@nigamkumar5646
6 жыл бұрын
Great sir thnk u so much its was look like very simple theorem in the way you explain it ...
@ArihantChawla
5 жыл бұрын
Atleast, like mention induction
@preetiyadav6260
Жыл бұрын
very helpful lecture
@myonlynick
7 жыл бұрын
0:42 I think the following statement is a bit more accurate to the one in the video ----->''if the set P* does not contain the empty word, then this solution is unique'' versus video's sentence which says: ''...has a unique solution...''
@rohitbale15
5 жыл бұрын
Whenever you apply Kleene closure(*) to any Regular Expression you will surely gonna get Empty word in the language set.
@AlinaMirzaCS-
4 жыл бұрын
what was that you used the same expression which we wanted to prove ... rip to this proof
@jethalalnhk2409
3 жыл бұрын
R = Q + RP is given to us we want to find solution to this and prove that R = QP*. Watch the whole video.
@marxman1010
3 жыл бұрын
@@jethalalnhk2409 The first step and second step are basically same, just replace R with QP* and show it works. But the uniqueness is not proved at all.
@jasindavid219
Ай бұрын
what is this sir question in question and soln in soln proof jai balaya
@user-su1pt5eu5e
2 жыл бұрын
sir whenever we take a string of length 'n' it would always greater in RP^n+1 so this term should be eliminated
@aydict
4 жыл бұрын
you almost had it, almost
@anastasistogkousidis9777
3 жыл бұрын
hahahahahahahahaha
@ishika6945
7 ай бұрын
at 2:34 how can R*R = R* ??? i think it should be R+ instead.
@divyanshudwivedi8452
3 жыл бұрын
can we write Arden's equation directly if regular grammar is given in question
@saibunny1253
Ай бұрын
I can see in the comment section about the nature of people. When sir did one single mistake everyone is putting laughing emojis . I don't know why. Didn't he help us through tough times ? No we just need to bash when someone makes mistake.
@h.raouzi175
6 жыл бұрын
how you can say , that you prooved it ??
@lone_wolf7721
3 жыл бұрын
I don't understand how can say p doesn't contain €(epsilon)
@zackcarl7861
2 жыл бұрын
We basically say ok ,that's my question that my solution , use them both to prove they are made for each other , you can use one to prove other 😆.
@AhamedKabeer-wn1jb
4 жыл бұрын
Thank you..
@rajivswargiary1536
7 жыл бұрын
I think you are following "Introduction to Automata and Compiler Design" book also by Dasaradh Ramaiah K. Publisher PHI
@Meri-bt9ry
7 ай бұрын
beautiful
@khushitripura3633
3 жыл бұрын
Thank you sir
@sjk9223
7 жыл бұрын
plz add videos of sequence detector, introduction to finite state model
@SaumyaSharma007
3 жыл бұрын
🤯 Sir as it is given that p doesn't contain epsilon so how can we write epsilon in the set of p i.e epsilon+p+p2+p3+...... to p* I think proof is using induction 😅 Kuch bhi chal rha h 😂
@gauravbhandari3457
3 жыл бұрын
Epsilon come from Q not p
@kavitimoulika5489
3 жыл бұрын
Excellent
@dhanushsivajaya1356
3 жыл бұрын
Thankyou sir
@17_jain_darsh65
3 жыл бұрын
interstellar's final scene is overrated in front of this
@garvitsingh2796
5 жыл бұрын
Do all lectures help us in gate exam also....all toc lectures
@maitreyakanitkar8742
3 жыл бұрын
2 minutes silence for the proof
@shanmugapriya7554
3 жыл бұрын
Y ardens theorem we need here?
@shashikalaraju5769
3 жыл бұрын
What is R here:(?
@chandiralekhats7135
Жыл бұрын
i have a doubt.what if P contains epsilon in R=Q+RP.
@mohammadzubairaalam3843
Жыл бұрын
then it will have infinite many solution
@sukamaldash3599
5 жыл бұрын
Huh!? What?! What just happened! . . . He proved it?! What!? O_O
@solarkadakiadam
7 жыл бұрын
You used the given solution to find the solution thats not how you prove a theorem
@MrPerfectpunk
7 жыл бұрын
That the point. He first proved that whether it is a solution to this equation or not. And in the next step he proved that whether this is the only solution or not.
@heranzhang6562
5 жыл бұрын
@@MrPerfectpunk So why can the second part proof the unique?
@nikhitawankhede2107
7 жыл бұрын
Sir what is reachable state and non reachable state
@shauryamukhopadhyay989
7 жыл бұрын
basically states that have nothing incoming into them are non reachable. States where you can reach to from the initial state (directly or via other states) is a reachable state. Check the last lecture on DFA minimization, it's explained there
@فيافيالتأملمهمةإصلاح
3 жыл бұрын
how tf can mathematicians demonstrate?like really it's impossible,u can't follow any method to get the answer either u already know similar ones or u can't at all 🤔
@dewanshkhandelwal5489
2 жыл бұрын
sir ji, I don't think this is a way to give a proof for the theorem you are using the result statement inside the proof to prove the same statement that is absolutely wrong :( How can you teach such wrong stuff to over youtube, please provide a genuine proof in the comment box
@raj-nq8ke
2 жыл бұрын
Obviously Proof is wrong. but good work on showing oberview.
@Pccoer_SECO
Ай бұрын
Used R=QP* to prove R=QP* 💀
@ekanshkumar7457
6 жыл бұрын
Thank u sir
@nishatsayyed8326
7 жыл бұрын
man do you use the mouse itself to write?? or any kind of stylus.... coz if you does it only with a fucking mouse.....then you should literally get an award for your writing skills.....👏👏✌️
@sofiyarao2063
7 жыл бұрын
ive been thinking on the same thing since the first lecture xD. This person is amazing no doubt..
@ajkdrag
7 жыл бұрын
stylus on tablet.
@gladyouseen8160
5 жыл бұрын
You. Didn't proved it man☹️☹️☹️
@kadambalapavan2280
3 жыл бұрын
we need to prove R=QP*, how can we use R=QP* in the proof without prooving it?. can anyone tell me
@marathi_manus467
3 жыл бұрын
Same doubt🤔🙄
@FortranCastle
8 ай бұрын
In proving an existential statement, you are allowed to assume any value for the thing whose existence you are trying to prove and then show that it satisfies the claim. Here we are just showing that R=QP* is a solution to the equation R = R + QP. All you need is to substitute QP* whenever R occurs on the RHS and see that it is a solution indeed. You are not proving that R equals QP*. You are proving that (R=QP*) is a solution to the other equation. It is the uniqueness part that did not work for me. I do not see how what he did proves uniqueness of the solution.
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