Guys!at 13:00 sir clearly mentioned that in AC->D if closure of AC and A is same then C becomes redundant. And also if we solve taking W->X it then holds transitivity with V->W , making the actual result V->X. Which directly comes by eliminating W rather than V.
Hello everyone, what is the correct procedure of finding redundancy on the LHS??? Actually in older video(3.8) and present video sir has taken different steps, in previous video: while checking if there's any redundancy on LHS at 11:35, sir has ignored one functional dependency which was "wz=>y"... But in the present video in the same method, at 19:45, sir has not ignored "vw=>x". And if we do not ignore "vw=>x", we get a different result.. What is the correct way!!!
According to what I understood and answer is turning to be correct: We check the LHS only after checking all the RHS so if an FD has survived the RHS round only then it can go for LHS round. In LHS round we check those FDs which have multiple attributes instead of single ones for redundancy. So in LHS we do not need to "ignore" the whole FD since we already did that in RHS. In short don't follow "ignore" method in RHS. In RHS just find closure vw with new set of FDS(means without y-->x since that got crossed in RHS round) that will be v,w,x Then find closure of v without any ignoring and same with w. Since v can give all the attributes same as vw so w is not needed.
In the second numerical you have told wrong while finding the redundancy in VW->X. When you were finding the closure of V u considered the Functional dependency VW->X so the answer is wrong. The correct answer is V->W, VW->X and Y->VZ.
this is correct, sir ne second numerical wrong solve kiya hain, woh bhul gye ki pehle wale numerical main unhone dependency ko ignore karke closure calcualte karna btaya hain.
in last lecture u said while calculating u ignore the term,here in 2nd question u didnt do tht...due to which v+=vwx,if u ignore the relation u wiill get v+=vw...making it non reduntant...
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While finding redundancy in left-hand side do we need to ignore or not....I mean, while checking for redundancy in (VW)--->X while finding V+ and W+ should we ignore (VW)--->X or not
while checking rhs in second ques while checking v closure you have also checked with vw->x but in previous video of canonical cover you said to ignore that fd in which we are checking....it's creating confusion
In the first example we get D-AC & AC-D , so by using transitivity can we write D-D, If we can write that does D-D signifies anything or it is just said to be trivial functional dependency?
Sir apne previous video me kaha tha ki closure same nikle par first wali f.d ko remove krte h... Or es video me apne bad wali f.d ko remove kia... Confusion... Please make me correct
as per this question agar previous vdo ka technique use kare toh answer me 'w' reh ja raha hai... thats wrong So ye wala method hi sahi hoga i guess... and this method works in the previous example too.
in the last step of the 2nd question v closure is same as vx closure and w ka closure is different . Then v should be redundant as it is same and should not be part of solution . Can u please clarify this ??. Thanks
Pratik Modi actually I have this doubt too, but then I understand it carefully than I found that the clousre of vw and v is same which is vwx. So w does not play any role there, because with w, we have the same clousre and without w , we also have the same clousre.
Pratik Modi what happened here is that for reducing the left hand side of dependency we find if any of the attribute among the given set of attribute can have same closure . Agar same hota hai then dusra wala attribute jo hai it is of no use kyuki mereko toh pehle attribute se hi mil jaa rha hai same value
Ab's la comedia see, A batsman took 2 balls to hit a 6, and a batsman took 1 ball to hit a six, then whom will you prefer to reduce the energy and time, offcourse the batsman with 1 ball. In the same way, VW and V has same clousre , but to reduce the redundancy we consider only V , not VW. Also, in Fxn Dependency, Logics in LHS and RHS are different , which you already learnt from that we can apply the Decomposition rule in RHS but not in LHS
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Guys!at 13:00 sir clearly mentioned that in AC->D if closure of AC and A is same then C becomes redundant.
And also if we solve taking W->X it then holds transitivity with V->W , making the actual result V->X.
Which directly comes by eliminating W rather than V.
whats up bro
Sir you are not less than Saint whose selfless efforts is for us
No one can teach like you sir...you teach in a very easy way...its great
Hello everyone, what is the correct procedure of finding redundancy on the LHS??? Actually in older video(3.8) and present video sir has taken different steps, in previous video: while checking if there's any redundancy on LHS at 11:35, sir has ignored one functional dependency which was "wz=>y"... But in the present video in the same method, at 19:45, sir has not ignored "vw=>x".
And if we do not ignore "vw=>x", we get a different result..
What is the correct way!!!
if you have got the answer for this question, kindly share with me also
@@abhijithkannan643 no reply from anyone yet bro🙁
Any solution??
According to what I understood and answer is turning to be correct:
We check the LHS only after checking all the RHS so if an FD has survived the RHS round only then it can go for LHS round. In LHS round we check those FDs which have multiple attributes instead of single ones for redundancy.
So in LHS we do not need to "ignore" the whole FD since we already did that in RHS. In short don't follow "ignore" method in RHS. In RHS just find closure vw with new set of FDS(means without y-->x since that got crossed in RHS round) that will be v,w,x
Then find closure of v without any ignoring and same with w. Since v can give all the attributes same as vw so w is not needed.
@@pastelteaaniiiiOkk.. Thanks!
There is problem sir, y+=yvxzw is right, Not y+=yvxzx
yes i also observe the same
Correct sir
Y>x should not be removed??
Yes sir ..he is right there is a mistake
Yes there is mistake there
wish if we could have these videos in our student life or in our 30's .
In the second numerical you have told wrong while finding the redundancy in VW->X. When you were finding the closure of V u considered the Functional dependency VW->X so the answer is wrong. The correct answer is V->W, VW->X and Y->VZ.
Please post your questions on our Official discussion forum t.me/KnowledgeGATE_forum
Yes.....same ans
I am noticed same problem please confirmed is your answer correct or not?
Confidently incorrect
this is correct, sir ne second numerical wrong solve kiya hain, woh bhul gye ki pehle wale numerical main unhone dependency ko ignore karke closure calcualte karna btaya hain.
Please upload video on this topic Lossless Join and Dependency Preserving Decomposition
in last lecture u said while calculating u ignore the term,here in 2nd question u didnt do tht...due to which v+=vwx,if u ignore the relation u wiill get v+=vw...making it non reduntant...
Same doubt
when you calculate the redundancy of left sight you do not need to ignore anything, sir said it in previous video
Thank you sir, You are 5th sem saviour for me. ❣️❣️ Even after 4 years your teching technique is so unique.....
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I think correct answer should be: V->W, W->X (because we removed V as it was redundant), Y->VZ
Yes, this is also what I am getting. Thanks
While finding redundancy in left-hand side do we need to ignore or not....I mean, while checking for redundancy in (VW)--->X while finding V+ and W+ should we ignore (VW)--->X or not
I also have the same doubt tomorrow is my exam, can u please tel weather we have to ignore it or not while checking?
if you had received the correct answer for this doubt kindly share
sir has taught this part wrong in both the videos
Thankkuu so much sir maza a gaya jis tarika s aap hame padhate ho and practice krba t ho 👏👏🖒
Minimal set of fd is affected by an order in which you are checking redundancy
while checking rhs in second ques while checking v closure you have also checked with vw->x but in previous video of canonical cover you said to ignore that fd in which we are checking....it's creating confusion
yes same question
Same question
same ques
Yes...how?
I also believe there is some confusion. The correct answer should be V->W, VW->X, Y->VZ
sir, devta ho aap 🙏🙏
Thanks sir aap bhot achcha pdate ho
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sir there is a problem at 18:07 y⁺=vxzw not y⁺=yvxzx and yₓ⁺=yvzwx
and by this what is the correct answer
plz reply
same ans
In the first example we get D-AC & AC-D , so by using transitivity can we write D-D, If we can write that does D-D signifies anything or it is just said to be trivial functional dependency?
It would be just trivial and wont make any sense.
Amazing....,sir plz upload videos on locking protocols as soon as possible
If to remove the redundancy on the LHS with five terms ABCDE->X,do we need to have 2^5-1 iteration
lot of respect from Gilgit Baltistan
Thank you.. keep learning !!
Awesome lecture 👍
Thank you sir 🙏❤️
Most welcome dear 😍Keep learning & supporting ! Do visit our website www.knowledgegate.in for more amazing courses & contents 👍😊
your videos helped me a lot sir 😊😊
Sir how to find out the number of different minimal cover possible..plz make a video related to this...
A->BC , B->AC ,C->AB will not work by this method . or we need to find specific sequence to implement this method . correct me if i am wrong.
sir ur videos are really awesome . can u plz make lectures on relational algebra.
Canonical cover for A-C,AC-D,E-AD,E-H
The answer of the last discussed question will be v->wx and y->v ....
b and b+ trees sir
Sir, if we cannot decompose lhs. Then why are we computing the closure of lhs and eliminating it? Indirectly we are decomposing itself right?
your teaching method and example clicks very well but my college professors not.....
Thanks sir ...you are great
namaskaar donston mein hun sanchittt.........My favourite lineeeeee
Great sir thank you so much
plzz upload B+ tree and locking protocols
Please tag the English videos. The ones where you at least explain in English as well as Hindi are very clear but it's hard to find them.
sir in last question redundancy is due to 'v' not 'w'.... please checkk
i think you are right. but i am still confused
love you sir , kya padhaya
Apse acha koi ni
You're a savior!
Thanks lot sir, This helped a lot in exam
C to b will also be redundant na cozzz we could derive d to b at that time
sir last q me aapne v closure nikala h at 19:52 usme x ni ayega therefore both v and x are essential
why u didn't ignore vw at the time of finding closure of v
Too good sir...👌👌👍👍
Thanks sir g
Sir, can I use vw->x rules in the closer of v+ 19:48 ? Please reply me ASAP..
Dear @Surajit, we're already using vw->x rule in the closer of v+. That's how we derived x.
just an hour left for my exam and watching this at 2x. Thankyou ❤
You're welcome 😊 All the best👍😊
13:15 why didnt you ignore AC->D sir?
Whats the reason?
In the past video for a similar case you ignored WZ->Y.
But why not now?
did you got the answer??
@@lalchandyogi1177 2 years man. I Don't remember now😂
very useful videos
Plz help me getting the minimal functional dependency F = {A -> BCE, AB -> DE, BI -> J}.
I think {A -> BCDE, BI -> J}
Hi can you please explain R(ABCDEFG) A--> B , ABCD-->EG,ACDF-->E ?
A-->B
ABCD-->EG
What is minimal set of functional dependencies
Thanks very much sir😆
21:25 why we eliminated W? not V?? did'nt get this point.
In closure of VW=VWX , what will be the the outcome if V=VWX AND W=VWX?
W
2:51 X ke hone na hone se kuch fark nahi pda, X extra hi thi. Hence proved XD
Sir apne previous video me kaha tha ki closure same nikle par first wali f.d ko remove krte h... Or es video me apne bad wali f.d ko remove kia... Confusion... Please make me correct
as per this question agar previous vdo ka technique use kare toh answer me 'w' reh ja raha hai... thats wrong
So ye wala method hi sahi hoga i guess... and this method works in the previous example too.
Sir, Can we apply other inference rules in the last step other than union rule if applicable?
Thanks sir but also upload software engineering artificial intelligence as UGC net point of view thanks again sir
LHS me kya krna hai smjh me nhi aya hai sir redundancy kaise hategi usme confusion horhi hai
Thank you sir
@Knowledge gate:. Sir, Did you upload videos for relational algebra?
I guess No
in the last step of the 2nd question v closure is same as vx closure and w ka closure is different . Then v should be redundant as it is same and should not be part of solution . Can u please clarify this ??. Thanks
Pratik Modi actually I have this doubt too, but then I understand it carefully than I found that the clousre of vw and v is same which is vwx. So w does not play any role there, because with w, we have the same clousre and without w , we also have the same clousre.
Pratik Modi what happened here is that for reducing the left hand side of dependency we find if any of the attribute among the given set of attribute can have same closure .
Agar same hota hai then dusra wala attribute jo hai it is of no use kyuki mereko toh pehle attribute se hi mil jaa rha hai same value
Ab's la comedia see, A batsman took 2 balls to hit a 6, and a batsman took 1 ball to hit a six, then whom will you prefer to reduce the energy and time, offcourse the batsman with 1 ball. In the same way, VW and V has same clousre , but to reduce the redundancy we consider only V , not VW.
Also, in Fxn Dependency, Logics in LHS and RHS are different , which you already learnt from that we can apply the Decomposition rule in RHS but not in LHS
Rajni Gupta I guess virtually we said same 😊
thnks alot both of uh😊
14:34 we have D -> AC AND AC -> D can't we cancel AC .
iam getting ans : V-->W, W-->X, Y-->VZ
Mistake at 18:01 pls check and correct
Sir either Complete us leave us....
wrong in second step...it should be y+->yxvwz
Thank you sir
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watch on 1.5x playback speed thank me later!
2X is even better
Y+ is written wrong sir
Board se thoda hatke khade hua kijiye sir
so at 14:32 ? AC--> D, and D --> AC, Can we not make one of them redundant? as they are basically the same dependencies but flipped the other way ?
your method would not work for finding minimal cover set of the following f.d.s {B -> A , D -> A, AB -> D}.
The answer should be B -> AD and D -> A
💕❤️❤️💕
while removing redundant function the only thing important was you
18:13 sir you are doing wrong.
Atleast try to reply your viewers .
You made a point in last step of second question which contradicts what you said in previous video
be grateful that he even makes the videos. he is a teacher. I suggest you be more respectful.
Did he speak English?
no, he spoke in hindi.
sir please explain in english
Bot sala
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Sir can we combine ac to bd ?
thank you sir
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