UHCL 26a Graduate Database Course - Preserving Dependencies

[@kacyraye]

You're a life saver. Spent hours looking at my class's slides, I should've just watched this 10 minute video.

[@a.pourihosseini]

very good algorithm. No such algorithm was explained in our own DB course, yet they expected us to determine if some composition was dependency preserving using just the definition ((Fx U Fy)+ = F+)

[@ryanwaldon1601]

Skip to 9:28, whole video summed up in 32 seconds lol

[@leKageBunshin]

+Ryan Waldon Thanks! XD

[@abir95571]

he made a wrong interpretation ..mostly the FD's are such that they are not contained in one relation completely,thus, trick is not a reliable option..it can used to reduce time

[@GaryBoetticher]

Sankhadip: You need to review the "trick" again. If the LHS and RHS of an FD is in some subschema, the FD is preserved. If the condition does not exist, then the FD needs to be checked using the long method. I do not claim that the "trick" will work for all FDs.

[@abir95571]

sir u misunderstood me...i was rather proposing that only when every attributes of an fd fits into a FD then only u can use the shortcut ,else the legitimate process has to be used..btw u did a preety impressive vid :)

[@sharik24]

Your examples, sir, are very clear and very well presented. Thank you very much for spending time making these videos.

[@bz967]

Thank you so much! This is very well explained. Just as said before, I should save my time watching this instead of going through slides for several times.

[@SpykarBro]

You are one of the best teachers ever ! Thanks a lot :)

[@christianllanas6349]

Well played sir you trolled me good. I appreciate you that you have taken the time to make these videos thank you.

[@fbakioglu]

Thank you very much for this video. I wish my professor was clear as much as you were. Very helpful.

[@yousrasaad7577]

Very useful lectures and an excellent teacher Thank you so much

[@DeepMajumder]

Sir, what would be the solution of this problem. R(A, B, C, D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition. a) A-> B, B->CD b) A->B, B->C, C->D c) AB -> C, C->AD d)A->BCD

[@eimaldorani]

What do you mean by key!? Candidate Key or super key!? Every Attribute is part of a super key, isn't it!?

[@sampathsurineni]

thank you very much. I didn't thought it is so easy.

[@razvanefros411]

I am very confused. is it not sufficient, for each FD, to check whether the attributes in the analyzed FD are all in at least one of the sub-schemas? like here with C->E we can clearly see that R2 contains both C and E therefore this FD is preserved, whereas for AB->C neither R1 nor R2 contain A,B, and C, therefore it is not preserved edit: i think i understand it better now. What i said before is almost sufficient. It can also be the case that the RHS of a FD is obtained in pieces from the 2 subschemas. For example, say we have A->BC in F, with R1 = {A,B} and R2 = {A,C}. R1 has the FD A->B and R2 has A->C. These 2 can be unioned into A->BC, which means the dependency is preserved. in other words even though B and C are not together in either of the subschemas, the dependency A->BC is preserved half in R1 and half in R2

[@sanayakhan6248]

just waaaaaaaaaaooooooooo.. thanx a lot 4 such awesome videos

[@eisfeleino5819]

HIGHLY APPRECIATE IT

[@Clamhead939]

Awesome man. Thanks! Very helpful.

[@priyasmit]

is it possible to get the notes you mentioned over internet sir?the videos series have been highly helpful

[@ayushgupta1089]

how you got R1 & R2? According to me R1(ABCE) & R2(BD).

[@malik4266]

Awsome Really helpful

[@floris6018]

Perfect! thanks for the effort

[@yogainalift]

I think there is a mistake in the video and I wonder why I am the only one seeing this. Gary is checking 2 times for the same Z=ABD in the same relation R2. Isnt this a mistake? Since you obviously will get the same result. Thanks for the video, Cheers

[@7dskei379]

iamthedood1 Yes, that is correct. It is easy to see that, however, the algorithm says to do it anyway because a computer would not be able to look ahead as you did.

[@fromthesanitarium]

you are just great sir I wish you were my professor

[@TejwanshSingh]

I wanted to ask that are there specialised ways to find out lossless decomposition and dependency preservation for 3NF and BCNF normal forms and is there any specialised algorithms for decomposition into 3nf and bcnf

[@TejwanshSingh]

One more thing you mention about notes in ur videos, are these notes available to us as well somewhere, can we also access these notes ??

[@GaryBoetticher]

I have videos on this.To decompose into 3NF: Bernstein's synthesis.To decompose into BCNF: Paired-attribute algorithm.

[@GaryBoetticher]

Yes, the there are notes available.You must register in my graduate DBMS class as a UHCL student.

[@TejwanshSingh]

+GaryBoetticher I am not a uhcl so I can't get them

[@Smartrixx]

You are using this same example to find lossy or lossless join decomposition. And it's lossless join which means we can find all the values using given decomposition. Based on that, can I say that every dependency is preserved?

[@Smartrixx]

Ohh ..... I got it....to find C using AB we must need to do a natural join. So yes dependency is not preserved but we can find the values. BTW nice videos....

[@Messi-rw9ng]

What's the name of this algorithm? Because so far I've just been naming it Gary's algorithm haha.

[@suprsg385]

well done understood it

[@vishnusbabu1]

thank you.

[@bksukriti4322]

thank you very much. it helped me a lot sir

[@JubayerRony]

thank you. the best lesson :)

[@nerosonic]

really helped, thanks!

[@kaustavbasu]

you saved me!!!

[@angus10292]

you rock!!

[@williamdewald9029]

What if Z intersect R is null? What is the closure of null?

[@GaryBoetticher]

Z has at least the LHS attributes of some FD.If Z intersect some subschema ends up in a null value, then Z does not change. Z grows monotonically, so it never loses any attributes.

[@williamdewald9029]

Thanks for the clarification.

[@alapatisrikanth3412]

how will you say A clouse is A ,C closure is ACE. i didn't get it

[@yooos3]

According to the given Functional dependencies, A alone doesn't determine any other attribute but itself. So A -> A is implied. Therefore, A+ = A But C can determine E and then E can determine A as C -> E, E -> A are given dependencies. Therefore C+ = CEA

[@yon1623]

what does Graduate Database Course mean? this is literally into DBMS.

[@ibrahimibrahim5225]

last 30 seconds == fire

[@camelCaseFTW]

nice :D

[@ThorkilKowalski]

You americans need a new word for the letter Z!

[@Messi-rw9ng]

zed

[@DeepMajumder]

Sir, what would be the solution of this problem. R(A, B, C, D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition. a) A-> B, B->CD b) A->B, B->C, C->D c) AB -> C, C->AD d)A->BCD