Lec 20: Problem on Lossless Join Decomposition in DBMS | Check whether Decomposition is Lossless

It is not dependency preserving

not necessarily candidate key.. it can be super key too

Thanks didi 🙂

you saved me, thanks a lot

Thank you man

If 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

Why 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

Thanks

Answer 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.

This relation is not Dependency preserving decomposition. Because F cannot derive A. IN the union of (R1 U R2 U R3).

Where did f go ?

Tqqq mam... 🤩🤩

Thnku so much mam for easy explanation

Can someone please tell me how to get the FDs of each decomposed relations in which mam has asked us to check Dependency Preserving condition

Great video

E is candidate key

AB-c is also

superb video mam

You always prove to be a life saver

I guess this video should come before #20 (the one preceding this). Thanks for all your amazing videos!

you are so beatiful, I like you

Ma'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

Your lecture are awesome ,thanks a lot mam

Sweet speed is 1.5x . Thank me later.

Super mam

Iobe u mam

Thank you mam <3

amazing video thank you so much

Ma'am I really appreciate your efforts 💗

Mam when are you going to make videos on multivalued ,4 th and 5th normal form??

Is 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...

C->D not covered so not dependency preserving

This is not DP.

not dependency preserving as AB->C and F->A are not present

MAM PLEASE UPLOAD 4TH AND 5TH NF

Damm short vedio and all concepts are perfectly clear... thankyou mam

Mam, 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

Your 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

mam if we have two common attributes in given decomposed subrelations then we should find closure for both attributes???

mam you are not fit for teaching gate subjects.

It 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😊.

Mam within how many days will u upload rest of the topics?

Not DP

Mam ap relational model or ER model par BHI viedo bna do please...

The given question is not dependency preserving decomposition. I am right or wrong please tell me mam... You are really good ... 👍❤️❤️

It's Not dependency preserving

Mam I have problem in few questions...... Can you please help me with that

PLEASE MAM CONTINUE THIS VEDIO LECTURES ON DBMS IT IS REALLY GOOD