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4. R1(WV) R2(WXYZ) W common hai, lekin W candidate key nhi hai. 1. R1(VWX) R2(XYZ) isme X common hai lekin X candidate key nhi hai, candidate key hone ke liye R2 me W hona chahiye na Candidate key: VW nikal rha hai
R1 aur R2 alg tables h aur inn dono tables m se kisi m bhi common attribute candidate key bn jati h toh lossless hoga (ik my reply to your comment is useless but for future viewer it may help)
Apne sir g meri bahut badi problem solve Kara Di ... Thank you sir g .... Amit sir toh pata nhi kya padhate h institute m .... Unhone aise padhaya iss method ko ... M toh usme ulajh Gaya really sir thank you very much
sir you are really great..the way you explain all those stuffs its really beautiful and i becomes your big fan... and i wanna say something that if possible then please upload videos on 4NF and 5NF. Thank You Sir.
exactly my question!!, up to the previous examples we calculated if the common attribute is the candidate key of the parent relation but in this question suddenly we are happy with the common attribute being the candidate key of either one of the child relations.
There's a mistake in the first part - Q6, wherein we have the three decompositions R1(a,b,c),R2(b,c,d) and R3(d,e). You say that the only common attribute between R1 and R2 is b, but it's actually b and c both!
Sanchit , can we try to solve this problem , Suppose that we decompose the relation R=(A,B,C,D,E,F) into three relations- R1(C, D, E) , R2(A, B, C) and R3(A, D, F) . Find whether this decomposition is lossless, if the following set F of functional dependencies holds. F={ A->BC; C->DE; D->F}
Hello sir, in the last option of table you calculated B as common between R1 and R2, but both of them has BC as common, still they satisfy 3rd condition?? Since C is not distinct. Please clarify.
we are not using B and C individually instead we are taking the composite key "BC", and as we know B is distinct so it does not matter if C is distinct or not because "BC" combined will always be distinct.
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sir i have a question :- since here we can find C.K with the help of functional dependency here so 1st we find the C.K and then check that the decomposed tables have those keys in common or not
No, we cant do that because if the common attribute is C.K for any of the decomposed tables then it will be lossless decomposition but it is not necessary ki wo common attribute CK prove hojaye using Functional dependencies like in last example X is not a C.K if you find it using functional dependencies but it is CK for R2 .
X is the candidate key for the second table. Only one condition is enough for it to prove that it is distinct he said so X isn't the candidate key of the first table.
and one more question:- does the child tables have to have the same C.K (a table can have multiple C.K )in common to satisfy the condition of loss-less decomposition
sauravjyoti kalita ........ you did not understand the question he raised ..........vw & xw are candidate keys for R(vwxyz) AND X is candidate key in R(xyz).....so x is candidatre key in R(xyz) and foreign key in R(vwx)!!understood?
sir in the 6th question R1(ABC),R2(BCD),R3(DE). If we combine R1 and R3 we get(ABCDE) but the intersection is null since there is nothing common in between them. So it should be Lossless decomposition.
Hello sir, in the last option of table of first question you calculated B as common between R1 and R2, but both of them has BC as common and C is not distinct, so it should be lossy.
Also , just have one question , whether we need to find candidate key for whole schema R and then we need to check whether this candidate key is available in decomposed schema R1 or R2 , or Whether we verify whether common attribute find by rule 2 , is candidate key in R2 or R2 ?
Just check if it is a candidate key for the particular relation R1 or not. If it is candidate key for the entire R, then it certainly would be for R1/R2.
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This lecture is "lossless" for sure sir. No time waste, straight to point perfect explanation!!
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This probably was the only lecture that i understood about "loseless join".........
I love the way u refer us as friends
are you an idiot??
@@sohammaity7389 simp Lmao
@@s7s7s7s7s7 best reply
4. R1(WV) R2(WXYZ)
W common hai, lekin W candidate key nhi hai.
1. R1(VWX) R2(XYZ) isme X common hai lekin X candidate key nhi hai, candidate key hone ke liye R2 me W hona chahiye na
Candidate key: VW nikal rha hai
R1 aur R2 alg tables h aur inn dono tables m se kisi m bhi common attribute candidate key bn jati h toh lossless hoga
(ik my reply to your comment is useless but for future viewer it may help)
Apne sir g meri bahut badi problem solve Kara Di ... Thank you sir g .... Amit sir toh pata nhi kya padhate h institute m .... Unhone aise padhaya iss method ko ... M toh usme ulajh Gaya really sir thank you very much
Sir literally maza aa raha is series me. 😊
Thank you so much Sir for such content!! ❤️🙏
You're most welcome dear 😍Keep learning & supporting ! Do visit our website www.knowledgegate.in for more amazing courses & contents 👍😊
We are blessed that....we got a tutor on youtube like you.....
Very nice explanation....thnku....☺
I hv no word sir...u r awsommm...thank u so much sir
sir you are really great..the way you explain all those stuffs its really beautiful and i becomes your big fan...
and i wanna say something that if possible then please upload videos on 4NF and 5NF.
Thank You Sir.
Hello sir, In second example 11:10 how is X candidate key? X alone cannot find W … Please correct me if I am wrong.
Correct
yes, also noticed the same
Guru g m toh issey bahut bade tareeke SE solve karta tha ..... Ab toh mouj aa gayi g
Wow sir g .... Dhansu trick h g ... Lossless , lossy ki
nice explantion sir.............Sir plz start uploading videos on computer org and architecture.........
sanchit sir, please upload video on dependency preserving as early as possible..
how x is candidate key when finding x closure w is missing in last question
proving candidate key in one relation is enough...........see previous video 3rd property!!!!!!!
i little bt confused and thanks
coommon attribute is a candidate key at least one of the table after orignal table divide itno two parts ...thanks ashish
Gautam sonu always welcome
exactly my question!!, up to the previous examples we calculated if the common attribute is the candidate key of the parent relation but in this question suddenly we are happy with the common attribute being the candidate key of either one of the child relations.
Thank you Sir for clear explaination.
Sir your teaching is fantabulous
This man is god 👍
There's a mistake in the first part - Q6, wherein we have the three decompositions R1(a,b,c),R2(b,c,d) and R3(d,e). You say that the only common attribute between R1 and R2 is b, but it's actually b and c both!
If B is distinct... then why we combine B with C
Same thought but idk why i am replying you its been almost 5 year for your watching this lecture.. 🙂😅
To kya dikkat hai bhai b and c dono common hai check (BC) key hai ya nahi or bo hai .....
Sanchit , can we try to solve this problem ,
Suppose that we decompose the relation R=(A,B,C,D,E,F) into three relations- R1(C, D, E) , R2(A, B, C) and R3(A, D, F) . Find whether this decomposition is lossless, if the following set F of functional dependencies holds.
F={ A->BC; C->DE; D->F}
mujhe achhe se samjh me aa gyi sir apka vedio
love your explanation
sir in 4th ques we r getting w as candidate key in R2.i think it is lossless.Plz kindly explain whether it is right or wrong
Hello sir, in the last option of table you calculated B as common between R1 and R2, but both of them has BC as common, still they satisfy 3rd condition?? Since C is not distinct. Please clarify.
take BC as key because R1 intersection R2=BC. He must have missed it.
we are not using B and C individually instead we are taking the composite key "BC", and as we know B is distinct so it does not matter if C is distinct or not because "BC" combined will always be distinct.
Very very thanks sir aur iski jarurat thi .
Thanku sir ji
Nice explanation
excellent explanation
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Sir ek bhi distinct hoga tab bhi 3rd property follow hogi?
Nice explanation sir.
🌹💞💞Awesome teaching sir 💞💞 🌹
Thank you sir.... great 🙏
Sir really very good explanation my each nd every doubt clear regarding this topic
Thankx a lot Sir!
Thanks 🙏 🙏
Sir acha explain kya apne..... Nyc
Thank you my friend.. Satah dete raho padhte raho.. so that main aur video bana pau aap logo ke liye..
Brilliant sir thanks
sanchit sir next please take on how to check fast dependency preservation .
Thnku sir ji...
Good job
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God bless you
Thanks a lot Gayatri..Stay blessed. Keep learning !!
Thank you
thank you sor🌹 🌹
sir i have a question :- since here we can find C.K with the help of functional dependency here so 1st we find the C.K and then check that the decomposed tables have those keys in common or not
No, we cant do that because if the common attribute is C.K for any of the decomposed tables then it will be lossless decomposition but it is not necessary ki wo common attribute CK prove hojaye using Functional dependencies like in last example X is not a C.K if you find it using functional dependencies but it is CK for R2 .
sir 6th mr R1 and R2 saare attribute cover nhi kr rhe h E nhi h dono me ...aise ke skte h kia h sir??
More questions on this topic will be helpful. Please upload soon if possible.
Do them by yourself
Sir last wale quetion mai x candidate key nai hai , to ye to lossi hua na
Great sir..
thank u
Thank you so much sir... Take a bow🙌
In the last example how you can fetch W by making X as a candidate key?
X is the candidate key for the second table.
Only one condition is enough for it to prove that it is distinct he said so X isn't the candidate key of the first table.
Thank you sir!
I have a confusion, here wv,wx,wy are the condidate keys.
Then 4th option should also be lossless. Please help
Thank u so much sir...
Thanks sir ji😆
Haat joda mat kijiye, we are blessed by you
Sir please...... upload the video of 4NF and5NF
🔥Complete DBMS by Sanchit Sir: tiny.cc/DBMS_Sanchitsir_kg
🔥🔥All Computer Science Subjects by Sanchit Sir: tiny.cc/CSbundle_dbms_kg
Love 😘you sir
Thanks a lot dear student.. keep learning and supporting !!
sir please upload videos on ER Model
Great
and one more question:- does the child tables have to have the same C.K (a table can have multiple C.K )in common to satisfy the condition of loss-less decomposition
bro foreign key reference hona chaiye bas
Sir i want more videos on dec omposition
How is X a candidate key? We cannot get W from X.
X is candidate key in relation R2
in r1 how it is possible
here i got the solution!!!!! common attribute X should be candidate key in either one of R1 & R2 (or) both.........refer previous video 3rd property!
Ashish Akash I think here X is only a part of candidate key....VW and XW are two candidate key
sauravjyoti kalita ........ you did not understand the question he raised ..........vw & xw are candidate keys for R(vwxyz) AND X is candidate key in R(xyz).....so x is candidatre key in R(xyz) and foreign key in R(vwx)!!understood?
sir please jaldi upload kijiega .
common attribute can be super key?? Anyone please
Sir thank you ❤
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opetion 4 is correct
May I know why 4th option is not crct for 2nd qn...? even it contains all attributes and vw ,it is also a candidate key.
W itself is not a candidate key
Sir kya agar ek bhi common attribute distinct ho to bhi vo lossless hoga?
sir in the 6th question R1(ABC),R2(BCD),R3(DE). If we combine R1 and R3 we get(ABCDE) but the intersection is null since there is nothing common in between them. So it should be Lossless decomposition.
maja aa gya
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Hello sir, in the last option of table of first question you calculated B as common between R1 and R2, but both of them has BC as common and C is not distinct, so it should be lossy.
yes brother
Ugc net Dec 2019 m tha ye questions
But Sir candidate keys ars vw and xw so how x only be a candidate key
Also , just have one question , whether we need to find candidate key for whole schema R and then we need to check whether this candidate key is available in decomposed schema R1 or R2 , or
Whether we verify whether common attribute find by rule 2 , is candidate key in R2 or R2 ?
Just check if it is a candidate key for the particular relation R1 or not. If it is candidate key for the entire R, then it certainly would be for R1/R2.
Yaar I need this reply as early as possible , can not wait for a month or many days.. plz help me
Last explaination was wrong.. please check sir, no lossless was found
sir i think this video u have given...i tried this on one sum bt couldn t get the answer...
if u ask me then i can send u in mail that sum if u want.....sir pls don t mind
100th comment❤️
Does anyone notices that he speaks very similar to technical guruji
"To fir chaliye suru kartehe" 🤣🤣
Inspired by Technical guruji
Maybe Technical Guruji was inspired by our Sanchit Guruji.. 🤣🤣
ii watch ur videos in 2x speed 😂
06/09/2024
you save my ass man !!!
Who's having exams tomorrow
bro does you test in practically! obviously no...😡
SIR YOU SPEAK TOO FAST !
Agar aap net k saare subject padhane lage toh .... M Bhagwan ki kasam kha kar kehta hu kamjor se kamjor bachha BHI net ko aik Baar m clear kar dega ..... Sachhi sir. .. m Kisi ki khaama khaa badaayi nhi karta hu ... Ap mein mujhe kuchh lagaa h tabhi m keh Raha hu g..
Why why english tittle when you talk on another language ??? BIG DISLIKE
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Thanks sir g
thanks
Thanks😊
Thanks a lot sir
Thank you so much, sir.
Thank you
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