CSDS - Computer & Data Science

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@actellimQT Ай бұрын

here from wikipedia!

@TE-uz5vg Ай бұрын

why didnt you prune 8 yet alpha was greater that beta at that instance but we pruned 5 instead

@ArvindLohar-h2k 3 ай бұрын

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

@saipavan5194 3 ай бұрын

Good❤❤

@freakninja1480 3 ай бұрын

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

@hassaanshakil 3 ай бұрын

Great video man! Loved the thorough explanation

@abbasnassar6114 7 ай бұрын

Thank you very much 🎃🎃

@Ragehunter 7 ай бұрын

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 🙏👌💯

@dokyungsoo8623 7 ай бұрын

Isnt closure of B is B,D,E?

@abuzerdag 7 ай бұрын

very clear and to the point, thank you very much

@singouini9687 8 ай бұрын

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

@JohnBogart-o9l 8 ай бұрын

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

@UTKARSHMISHRA-rc8lt 9 ай бұрын

one of the best videos out there on BCNF

@vrdrivesolutions3695 9 ай бұрын

Great vid, easy to understand

@danielmooney8719 10 ай бұрын

Great explanation dude!

@trey_tipton 10 ай бұрын

This made creating NFA's so much easier, thank you so much!

@omrifishbein7159 10 ай бұрын

Thank you. You forgot to explicitly mention that pruning means an alpha beta state in a node that "signals" it to stop investigating its remaining children. You also forgot to return the value of the right child of the root to the root and then reject it... Your stress on first updating (yes/no) v and only after that updating (yes/no) alpha or beta is nice. Also, indicating that alpha/beta values are passed down wards and v values are passed upwards is nice. Perhaps it is worth adding to this "general view" of "information passing" that as v values are passed upwards, alpha beta values may update...

@valerianafuna8972 Жыл бұрын

this is the only video that i understood

@spamcolector Жыл бұрын

Thank you, very useful for CS students.

@sumithreddy1088 Жыл бұрын

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

@HayotbekKamchiev-w4q Жыл бұрын

Thank you for your time and dedication. You made it pretty simple to understand

@Dbanj598 Жыл бұрын

Good explanation!

@norhanecherif Жыл бұрын

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

@diegomorales4843 Жыл бұрын

Wow, amazing explanation, thank you so much!

@taliamoshe8578 Жыл бұрын

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

@JeepartN Жыл бұрын

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

@csds-computerdatascience3452 Жыл бұрын

{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

@muhammadsameer1512 2 жыл бұрын

Thanks Sir...

@theguitartist__ 2 жыл бұрын

life saving stuff, thank you so much

@XD-wn6li 2 жыл бұрын

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

@csds-computerdatascience3452 2 жыл бұрын

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.

@Michael-wh4my 2 жыл бұрын

@988skywalker 2 жыл бұрын

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 2 жыл бұрын

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 2 жыл бұрын

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

@jl-wi4oh 2 жыл бұрын

legend

@aryankhandelwal1659 2 жыл бұрын

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

@Νεόπλαστος 2 жыл бұрын

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 2 жыл бұрын

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

@Νεόπλαστος 2 жыл бұрын

@@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 Жыл бұрын

@@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.

@bamusalako 2 жыл бұрын

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

@platinoob__2495 2 жыл бұрын

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 2 жыл бұрын

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 2 жыл бұрын

@@csds-computerdatascience3452 thank you very much