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It is not dependency preserving
Ответитьnot necessarily candidate key.. it can be super key too
ОтветитьThanks didi 🙂
Ответитьyou saved me, thanks a lot
ОтветитьThank you man
ОтветитьIf you take first R2 and R3 then R23 is satisfy all three condition then combine with R1 and R23 all three condition is satisfy so i think this is lossless decomposition
ОтветитьWhy all are saying F->A not covered?,F->A is there right? Because in dependency EF,when finding F closure we get F->A right? Then how F->A is not there?i cant understand
ОтветитьThanks
ОтветитьAnswer for the last question: I got { C --> AB, AB ---> C, C --> DE, D --> EC, E --> F }. Comparing with the original FDs, the one I found does not contain F --> A, either directly or indirectly.
ОтветитьThis relation is not Dependency preserving decomposition. Because F cannot derive A. IN the union of (R1 U R2 U R3).
ОтветитьWhere did f go ?
ОтветитьTqqq mam... 🤩🤩
ОтветитьThnku so much mam for easy explanation
ОтветитьCan someone please tell me how to get the FDs of each decomposed relations in which mam has asked us to check Dependency Preserving condition
ОтветитьGreat video
ОтветитьE is candidate key
ОтветитьAB-c is also
Ответитьsuperb video mam
ОтветитьI guess this video should come before #20 (the one preceding this). Thanks for all your amazing videos!
Ответитьyou are so beatiful, I like you
ОтветитьMa'am You save Exam.....Ma:'am if you don't mind may I ask you something....?
Ma'am are you already engaged?......If not
Aapse sadi karke life time aapse sikhna chahata hoon.....
Not dp as F->A is not present in the union of all the sub relations
ОтветитьYour lecture are awesome ,thanks a lot mam
ОтветитьSweet speed is 1.5x . Thank me later.
ОтветитьSuper mam
ОтветитьIobe u mam
ОтветитьThank you mam <3
Ответитьamazing video thank you so much
ОтветитьMa'am I really appreciate your efforts 💗
ОтветитьMam when are you going to make videos on multivalued ,4 th and 5th normal form??
ОтветитьIs the common attribute should be candidate key or it can be super key ...?cause in previous videos at some point you told it was super key and after that you are saying for candidate key all the time...
ОтветитьC->D not covered so not dependency preserving
ОтветитьThis is not DP.
Ответитьnot dependency preserving as AB->C and F->A are not present
ОтветитьMAM PLEASE UPLOAD 4TH AND 5TH NF
ОтветитьDamm short vedio and all concepts are perfectly clear... thankyou mam
ОтветитьMam, thank you for such a wonderful session. I have got the ans for your last qsn as ‘ decomposition is not dependency preserving’ is it correct ?request you to please share the ans
ОтветитьYour videos are great! Thank you, Small question. Can the common attribute in a two relations be a composite key? Lets say the Composite key is the candidate key in one relation. can we say that it is Lossless? Thank you
Ответитьmam if we have two common attributes in given decomposed subrelations then we should find closure for both attributes???
Ответитьmam you are not fit for teaching gate subjects.
ОтветитьIt is not dependency preserving as F can't determine A.
Even at 1.5x speed I can understand her very easily and saves a lot of time. Thank you ma'am for your videos these are well explained.
Mam, the growth of your channel itself shows how much students love your channel😊.
ОтветитьMam within how many days will u upload rest of the topics?
ОтветитьNot DP
ОтветитьMam ap relational model or ER model par BHI viedo bna do please...
ОтветитьThe given question is not dependency preserving decomposition. I am right or wrong please tell me mam... You are really good ... 👍❤️❤️
ОтветитьIt's Not dependency preserving
ОтветитьMam I have problem in few questions...... Can you please help me with that
ОтветитьPLEASE MAM CONTINUE THIS VEDIO LECTURES ON DBMS IT IS REALLY GOOD
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