BCNF Decomposition

[@JohnBogart-o9l]

Amazing, you managed to make a clear explanation in 8 minutes whereas my professor confused the entire class over 2 lectures

[@syedrafayullah6749]

Thank you so much, this is the only video that helped me bro.. I have been trying to understand this all semester...

[@thierrybeaulieu4403]

You're now partially responsable of my degree

[@UTKARSHMISHRA-rc8lt]

one of the best videos out there on BCNF

[@singouini9687]

Life saver, out of the pool of videos, yours is the only one that did the job! Thanks

[@Ragehunter]

No cap, after searching and listening all BCNF videos on KZbin for DBMS finals this video is BEST VIDEO for BCNF. If you see this please dont skip this gem video. Thank you so much man, amazing explonation 🙏👌💯

[@spamcolector]

Thank you, very useful for CS students.

[@valerianafuna8972]

this is the only video that i understood

[@bamusalako]

gj that was very well explained and good sheets instandly understood, damn its hard to find good explaining videos but when youve found one youre happy and i am now ty man

[@sumithreddy1088]

Thanks a lot !!!!! Was confused with this topic but ur lecture is very clear !!!!!!

[@aryankhandelwal1659]

nicely covered all concept of bcnf decomposition in such a short video great man👍👍

[@norhanecherif]

thank you so much , this is the only video that helped me

[@theguitartist__]

life saving stuff, thank you so much

[@abbasnassar6114]

Thank you very much 🎃🎃

[@diegomorales4843]

Wow, amazing explanation, thank you so much!

[@Νεόπλαστος]

FDs A->C and B->D have attributes that depend on part of the candidate key, so this relationship is not even in 2NF, let alone 3NF. So, how can you proceed with BCNF decomposition if it is not 3NF?

[@martinlunn9660]

The relation doesn't need to be in any specific form to be decomposed.

[@Νεόπλαστος]

@@martinlunn9660 I think you are mistaken. BCNF is a subset of 3NF which in turn is a subset of 2NF. To decompose to BCNF, the FDs need to be in 3NF. Do not forget that FDs are describing a database schema so "relation doesn't need to be in any specific form to be decomposed' is invalid since you are not decomposing relations, but rather database schemas.

[@abdullahshahid9051]

@@Νεόπλαστος Strictly speaking, a relation doesn't need to be in 2NF to apply BCNF decomposition. However, it is more common to first ensure that a relation is in 2NF before moving on to decompose it into BCNF, as the process of normalization typically follows a step-by-step progression from 1NF to 2NF to 3NF and then to BCNF. If you decompose a relation directly into BCNF, it will automatically satisfy the requirements of 2NF (and 3NF) as well. This is because BCNF is a stronger normal form than 2NF and 3NF.

[@bhavyakumar7423]

Very well done! Keep it up!!

[@dokyungsoo8623]

Isnt closure of B is B,D,E?

[@988skywalker]

For Canditate Keys for something like R4 lets say you end up with something like (A, C, F, G, E) would that be ACF->GE or AC->FGE or would it be ACFG->E, I was a bit of confused did this arrow just need to be true for the functional dependencies given in the problem I was working on?

[@csds-computerdatascience3452]

Candidate keys are defined by the known functional dependencies. The candidate key for (A,C,F,G,E) would be the minimum amount of attributes that there closure has all the attributes of the relation (A,C,F,G,E)

[@988skywalker]

@@csds-computerdatascience3452 oh okay is it necessary to find all super keys at the beginning or is one sufficient

[@platinoob__2495]

If i have, for example, R(A,B,C,D,E,F) and F is not in any Functional dependency, what do I do?

[@csds-computerdatascience3452]

In this case, we use the functional dependency F->F, because F can be determined from itself. The rest of the decomposition is according to the steps in the video.

[@platinoob__2495]

@@csds-computerdatascience3452 thank you very much

[@actellimQT]

here from wikipedia!

[@jl-wi4oh]

legend

[@XD-wn6li]

why do you show AB->AB in R6? don't we want to show AB->CD?

[@csds-computerdatascience3452]

C & D are not in the partial relation R6, therefore the relation AB-> CD is not present in the relation R6. In each partial relation we are looking for the relevant functional dependencies, and in R6 this is only AB->AB.

[@JeepartN]

im sorry but what does the plus mean in {D}⁺ ?

[@csds-computerdatascience3452]

{D}⁺- is the closure of the attribute D. The plus sign refers to the closure. If the term closure isn't familiar, I suggest checking out my video on the closure of an attribute. kzbin.info/www/bejne/iIiUaJ2jbs2Fatk

[@anishmanandhar1203]

Alot of videos I see has refered R - X for X-> A and XA are you sure that this is correct?

[@taliamoshe8578]

wait why you havnt put the R=(ABCD) from the start? you rather smaller R's?

[@muhammadsameer1512]

Thanks Sir...

[@ArvindLohar-h2k]

why we should create table of AB->AB .. whyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

[@freakninja1480]

Thats why this video only got 800likes, that video is full of crap, so confusing and just a waste of time, dont watch it guys

[@Michael-wh4my]

[@Exquisite_Blue]

My guy it just does not compute in my brain. F

[@csds-computerdatascience3452]

Sorry to hear that my friend