Lec-37: Dependency Preserving Decomposition in DBMS | Example 2 in Hindi

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In the dependency preservation, at least one decomposed table must satisfy every dependency. If a relation R is decomposed into relation R1 and R2, then the dependencies of R either must be a part of R1 or R2 or must be derivable from the combination of functional dependencies of R1 and R2.
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[@muhammadikram139]

Love 💗 and respect from 🇵🇰
In simple words firstly we check the validity of Functional dependencies (FDs) of decomposed tables matching them all FDs with original FDs on top to find the real and valid FDs of decomposed tables and then secondly we map the FDs of original table down to decomposed table FDs to determed wether they are preserved or not.

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[@tejaswinibs8840]

Very well done.....super explanation🙂

[@criminalcoder6307]

This is some explanation....
Varun sir

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I'm learning from your lectures and solving competitive problems very easily.

[@logic738]

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[@vaishnavisaxena4699]

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[@Hihihi190]

Dhanywaad 😊

[@ranuyadav7913]

Thank you sir 🙂

[@Hrit]

This example was very good.

[@apuu7476]

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[@panditsanjay7762]

Sir ye same question net m aaya tha good explanation

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[@acmsociety7739]

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[@vinitasingh1117]

Thank you sir

[@prekshajain3980]

Thank you so much Sir aap bhot bhot achche teacher hain aapki vajah se subject se dar nhi lgta😊

[@rahulmaurya5236]

Very nice explanation

[@UMANGTAILANG28]

One doubt, while determining the FDs for R2 table (BCD), why did we not consider the simpler FDs like B -> C, C -> B, C -> D, D-> C, B -> D and D -> B?

[@niharkajla8234]

As per decomposition property, B->CD is equivalent to B->C and B->D. Hence cases mentioned by you are covered.

[@honeyverma2755]

@@niharkajla8234 thank you bro for help

[@ra.jee.v]

@@niharkajla8234 but if b->c is true and b->d is false then it will give wrong answer n?

[@Hrit]

@@niharkajla8234 Thanks

[@satyadharkumarchintagunta3793]

@@niharkajla8234 Thank You bro for the help

[@yashparmar5722]

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[@frontendEngineer-ne2sn]

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[@minivashishta9288]

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danyavad

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[@BiddyutHasan]

very good

[@GyanaranjanPatiMCA]

Thx sir

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[@ananyasingh7696]

Sir please make tutorials for engineering maths as well

[@GyanaranjanPatiMCA]

Select count(thx) as for_sir from student
--ans
--100

[@animeshhazra6061]

Nice

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[@CMPER-zg1yv]

Is this a lossless or lossy decomposition ?

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[@GateSmashers]

Sure...

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[@prashunprateek9085]

[@seemadahiya432]

Sir is this the only method to solve this ques..... Or we can solve. It by key concept like in R1 and R2 D is common so D should be key in any of the decomposed relation sir plz reply I m right or wrong

[@asitkumarsahu5211]

@seema dahiya But in the above example after decomposed into two table D is common .. Have u achieved dependency preserving decomposition...

[@jasteenkaur7315]

where can i find practice questions of this topic?

[@obviousness251]

Sir in this question, isn't it given that D->A and BD->C, so won't it be BA=BD, and then BA->C, so we will achieve the original FD

[@obviousness251]

Please help

[@arpitjain6121]

Same question
Please help sir

[@Man_of_Culture.]

For Pseudo Transitivity there should X->Y and WY->Z then we can say WX->Z. I hope its helpful.

[@mvenkataadityaprakash9485]

Sir, at 4:57 (BC) will also satisfy from BC->D

[@30sunique78]

Check consciously , there only One error that is CD is CD->{c ,d ,b a } there sir not consider this I don't know why

[@varshabhosale4716]

sir can you please tell if given decomposition is dependency preserving or not. X(PQRS) FD= { QR -> S, R -> P, S -> Q} decomposed to Y(PR) AND Z(QRS) this question is from gate 2019

[@suchismitadalal1693]

No its not preserving.S->Q is not preserving.

[@aloksinghrathore2791]

@@suchismitadalal1693 actually S->Q is also preserving if we see carefully. So dependencies are preserved in this question.

[@meghabagri7370]

@@aloksinghrathore2791 No S -> Q is not preserving.

[@aloksinghrathore2791]

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[@tayyabsandhu8066]

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[@abishekbabu9247]

sir ji why you did not took like B->C,B->D...... individually, would you please explain.

[@harshshah9962]

Bro what does B->CD means, we can write it as B->C, B->D that's why he didn't write it down

[@dustudoyitto]

if D -> A and AB-> CD so DB -> CD . Is it not possible ??

[@GeetChhabra-hx5gt]

mera bhi ye hi doubt hai bhai ? so did he make a mistake or you cant do it his way😭?

[@jenilshah2231]

@@GeetChhabra-hx5gt you can do it this way infact AB->CD also means A->C and B->D

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