@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.
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
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...''
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
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*
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*
@@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.
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
@@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".
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.
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.
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
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 🤔
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
🤯 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 😂
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.....👏👏✌️
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.
inception movie is over rated in front of arden's theorem's proof
@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.
@@sohamshinde1258share they website link to me
@@chandhunukala6949 He removed that commment I didnt had it saved :(
🤣🤣🤣
I used the stones to destroy the stones😂
lmaooo
Yees he proved it with what he has to prove.
Worst
mann that is really true 😂
😂😂 marvel fan everywhere... Thanos was right 😂
2 min of silence ....what a proof
same here bruhhhhhhhhh
`I used the R = QP* to derive the R = QP*`
Idiotic proof
@@cyanfroste5559 oo i see a man of culture
@@cyanfroste5559 Derive the uniqueness, but why it means unique answer?
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
Love ur teaching bro❤
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...''
Whenever you apply Kleene closure(*) to any Regular Expression you will surely gonna get Empty word in the language set.
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
thankss broo
✨👑
R kaha gya daya kuch to gadbad hai
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*
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*
Sometimes sir's Genius… It's Almost Frightening.
XD
It's funny what he did but that one thing makes me remember Arden's Theorm forever
his genius has a gravity of its own
How to fire a fire take out the fire from fire😂😂😂
you funny person
Amazing explanation ❤
I KNEW NESO IS THE BEST EVER TUTOR
what was that you used the same expression which we wanted to prove ... rip to this proof
R = Q + RP is given to us we want to find solution to this and prove that R = QP*. Watch the whole video.
@@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.
Mindblowing Explain ability of Nesco academy.Thank you sir. ❤
😂
Thank you sir.and sir please post some problem on Arden theorem and how to convert a regular expression into finite automata.please sir .
very helpful lecture
Atleast, like mention induction
you almost had it, almost
hahahahahahahahaha
Thank you very much sir
Good explanation
Great sir thnk u so much its was look like very simple theorem in the way you explain it ...
at 2:34 how can R*R = R* ??? i think it should be R+ instead.
how you can say , that you prooved it ??
can we write Arden's equation directly if regular grammar is given in question
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
And what about R inside the expression where it is gone
@@dailymemes2512 he put the value of r agin and again
this is right. In the video, that guy is proving using the statement that is to be proved.
Much better. I can sleep in peace now
@@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".
Thank you sir
2 minutes silence for the proof
Thank you..
You used the given solution to find the solution thats not how you prove a theorem
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.
@@MrPerfectpunk So why can the second part proof the unique?
I don't understand how can say p doesn't contain €(epsilon)
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 😆.
Thankyou sir
sir whenever we take a string of length 'n' it would always greater in RP^n+1 so this term should be eliminated
plz add videos of sequence detector, introduction to finite state model
Excellent
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.
I think you are following "Introduction to Automata and Compiler Design" book also by Dasaradh Ramaiah K. Publisher PHI
what is this sir
question in question and soln in soln proof
jai balaya
Do all lectures help us in gate exam also....all toc lectures
interstellar's final scene is overrated in front of this
beautiful
Y ardens theorem we need here?
What is R here:(?
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
i have a doubt.what if P contains epsilon in R=Q+RP.
then it will have infinite many solution
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 🤔
Sir what is reachable state and non reachable state
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
You. Didn't proved it man☹️☹️☹️
Huh!? What?! What just happened!
.
.
.
He proved it?! What!? O_O
🤯
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 😂
Epsilon come from Q not p
Thank u sir
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.....👏👏✌️
ive been thinking on the same thing since the first lecture xD. This person is amazing no doubt..
stylus on tablet.
Used R=QP* to prove R=QP* 💀
O bhai..
bhai e proff hai kya 🤣🤣🤣
Obviously Proof is wrong. but good work on showing oberview.
Kya kiye ho, aisa proof dekh-ke neend nahi aayegi ab.. 🙄
r u from assam?
yes
@@parikshit804 chutiye tereko nhi pucha usne.
we need to prove R=QP*, how can we use R=QP* in the proof without prooving it?. can anyone tell me
Same doubt🤔🙄
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.
❤️
Damn
op
awesome
first dislike from my side on this channel
Ye to ratta marna pdega
To much of hitch-potch. Make it clear what you try to say. So much of amateurish attitude. Too bad .
then learn it yourself, why even are you online, learn from book if you're so mature
Thank you..