UHCL 26a Graduate Database Course - Preserving Dependencies

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GaryBoetticher

GaryBoetticher

Күн бұрын

Пікірлер: 53
@kacyraye
@kacyraye 10 жыл бұрын
You're a life saver. Spent hours looking at my class's slides, I should've just watched this 10 minute video.
@a.pourihosseini
@a.pourihosseini 5 жыл бұрын
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
@ryanwaldon1601 9 жыл бұрын
Skip to 9:28, whole video summed up in 32 seconds lol
@leKageBunshin
@leKageBunshin 8 жыл бұрын
+Ryan Waldon Thanks! XD
@abir95571
@abir95571 8 жыл бұрын
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
@GaryBoetticher 8 жыл бұрын
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
@abir95571 8 жыл бұрын
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
@sharik24 13 жыл бұрын
Your examples, sir, are very clear and very well presented. Thank you very much for spending time making these videos.
@bz967
@bz967 10 жыл бұрын
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
@SpykarBro 10 жыл бұрын
You are one of the best teachers ever ! Thanks a lot :)
@christianllanas6349
@christianllanas6349 7 жыл бұрын
Well played sir you trolled me good. I appreciate you that you have taken the time to make these videos thank you.
@fbakioglu
@fbakioglu 12 жыл бұрын
Thank you very much for this video. I wish my professor was clear as much as you were. Very helpful.
@yousrasaad7577
@yousrasaad7577 10 жыл бұрын
Very useful lectures and an excellent teacher Thank you so much
@DeepMajumder
@DeepMajumder 9 жыл бұрын
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
@eimaldorani 7 жыл бұрын
What do you mean by key!? Candidate Key or super key!? Every Attribute is part of a super key, isn't it!?
@sampathsurineni
@sampathsurineni 11 жыл бұрын
thank you very much. I didn't thought it is so easy.
@razvanefros411
@razvanefros411 11 ай бұрын
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
@sanayakhan6248 10 жыл бұрын
just waaaaaaaaaaooooooooo.. thanx a lot 4 such awesome videos
@eisfeleino5819
@eisfeleino5819 7 жыл бұрын
HIGHLY APPRECIATE IT
@Clamhead939
@Clamhead939 12 жыл бұрын
Awesome man. Thanks! Very helpful.
@priyasmit
@priyasmit 13 жыл бұрын
is it possible to get the notes you mentioned over internet sir?the videos series have been highly helpful
@ayushgupta1089
@ayushgupta1089 6 жыл бұрын
how you got R1 & R2? According to me R1(ABCE) & R2(BD).
@malik4266
@malik4266 10 жыл бұрын
Awsome Really helpful
@floris6018
@floris6018 10 жыл бұрын
Perfect! thanks for the effort
@yogainalift
@yogainalift 9 жыл бұрын
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
@7dskei379 9 жыл бұрын
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
@fromthesanitarium 11 жыл бұрын
you are just great sir I wish you were my professor
@TejwanshSingh
@TejwanshSingh 9 жыл бұрын
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
@TejwanshSingh 9 жыл бұрын
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
@GaryBoetticher 9 жыл бұрын
I have videos on this.To decompose into 3NF: Bernstein's synthesis.To decompose into BCNF: Paired-attribute algorithm.
@GaryBoetticher
@GaryBoetticher 9 жыл бұрын
Yes, the there are notes available.You must register in my graduate DBMS class as a UHCL student.
@TejwanshSingh
@TejwanshSingh 9 жыл бұрын
+GaryBoetticher I am not a uhcl so I can't get them
@Smartrixx
@Smartrixx 7 жыл бұрын
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
@Smartrixx 7 жыл бұрын
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
@Messi-rw9ng 2 ай бұрын
What's the name of this algorithm? Because so far I've just been naming it Gary's algorithm haha.
@suprsg385
@suprsg385 8 жыл бұрын
well done understood it
@vishnusbabu1
@vishnusbabu1 10 жыл бұрын
thank you.
@bksukriti4322
@bksukriti4322 11 жыл бұрын
thank you very much. it helped me a lot sir
@JubayerRony
@JubayerRony 10 жыл бұрын
thank you. the best lesson :)
@nerosonic
@nerosonic 11 жыл бұрын
really helped, thanks!
@kaustavbasu
@kaustavbasu 6 жыл бұрын
you saved me!!!
@angus10292
@angus10292 9 жыл бұрын
you rock!!
@williamdewald9029
@williamdewald9029 9 жыл бұрын
What if Z intersect R is null? What is the closure of null?
@GaryBoetticher
@GaryBoetticher 9 жыл бұрын
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
@williamdewald9029 9 жыл бұрын
Thanks for the clarification.
@alapatisrikanth3412
@alapatisrikanth3412 8 жыл бұрын
how will you say A clouse is A ,C closure is ACE. i didn't get it
@yooos3
@yooos3 8 жыл бұрын
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
@yon1623 Жыл бұрын
what does Graduate Database Course mean? this is literally into DBMS.
@ibrahimibrahim5225
@ibrahimibrahim5225 7 жыл бұрын
last 30 seconds == fire
@camelCaseFTW
@camelCaseFTW 12 жыл бұрын
nice :D
@ThorkilKowalski
@ThorkilKowalski 9 жыл бұрын
You americans need a new word for the letter Z!
@Messi-rw9ng
@Messi-rw9ng 2 ай бұрын
zed
@DeepMajumder
@DeepMajumder 9 жыл бұрын
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
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