what does it mean 6:35 that "C" was las thing that was included? Included into what and where? having some trouble understanding that part. Thank you gary from australia
Initially he starts with (R - AD), which means 'A' was included from the start. Then he adds to that 'BC' (following the calculation of R - AD). He then removes 'D', then 'B' and are left with 'C' which is the element not included originally in the (R - AD).
He creates a variable Y that contains all the attributes of C then he takes the R-AD again and gets the BC closure. Since the closure contains both A and D he can remove one of them and he chooses to remove D. As you can see Y = ABC, which means C closure is C.
I have a query sir, we know that a given F is in 2NF if there is no 2nf voilation, but how do we check if there is any 3NF voilation as well and then when we use bernsteins algorithm we get to 3nf form and from there how do we check if there is any BCNF voilation. i mean what exactly is the difference between a super key and candidate key ??? For Eg if the candidate keys are ABC and ABD then would the super key be ABCD and the non prime attributes can also be refered as non key attributes ?
I'm having a lot of difficulty here. Why does dropping B from (R - BC) give you AC? R - C = ABD, and ABD+ = ABCD, right? And what do you mean by "the last thing that was included?"
Hi Gary, i was wondering if you deleted one of your videos that showed decomposition of functional dependency for BCNF without using the paired attribute algorithm. Can't seem to find it. I'll appreciate your reply.
+Gary Boetticher Thank you for the reply, another pressing question I have is lets suppose I have the set of functional dependencies as follows: BG->CD; G->F; CD->GH; C->FG F->D I know for the fact that by deploying your LMR method as described in previous videos that the key can be either BG or BE, applying the algorithm described in the video I obtain the following decomposition G(G,F,D) and Rest(B,C,H) my question to you is, can I decompose these relations any further?
+Gilbert Adu-Sefah Yes, I think it is. Because when R - BC = A+ = ACD (he included the D here. From the same logic C+ should have been CD). However even if C+ = CD, I do not think that this has an effect on the overall result in this problem.
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Thanks for this Dr. Gary, I learned more in 30 mins of watching your videos than I did whole term from my professor.
YOU ARE LITERALLY THE MAN
My teacher said we have to know how to decompose into BCNF, yet didn't teach us how. Thank u!!! I'm just in undergrad but this was so helpful
what does it mean 6:35 that "C" was las thing that was included?
Included into what and where? having some trouble understanding that part.
Thank you gary from australia
Initially he starts with (R - AD), which means 'A' was included from the start. Then he adds to that 'BC' (following the calculation of R - AD). He then removes 'D', then 'B' and are left with 'C' which is the element not included originally in the (R - AD).
He creates a variable Y that contains all the attributes of C then he takes the R-AD again and gets the BC closure. Since the closure contains both A and D he can remove one of them and he chooses to remove D. As you can see Y = ABC, which means C closure is C.
I have a query sir,
we know that a given F is in 2NF if there is no 2nf voilation, but how do we check if there is any 3NF voilation as well and then when we use bernsteins algorithm we get to 3nf form and from there how do we check if there is any BCNF voilation.
i mean what exactly is the difference between a super key and candidate key ???
For Eg if the candidate keys are ABC and ABD
then would the super key be ABCD
and the non prime attributes can also be refered as non key attributes ?
What i cant understand es why did you stop in (R - AD) = BC+? why not continuin with all the possible combinations?
Sir can you please let us know that in the case:
when we have R(A,B,C,D,E)
and F(A->B, C->D)
then for (R-AD)=BCE+=?
I'm having a lot of difficulty here. Why does dropping B from (R - BC) give you AC? R - C = ABD, and ABD+ = ABCD, right? And what do you mean by "the last thing that was included?"
Hi Gary, i was wondering if you deleted one of your videos that showed decomposition of functional dependency for BCNF without using the paired attribute algorithm. Can't seem to find it. I'll appreciate your reply.
+Paschal Oduoza Please go to my course website at sce.uhcl.edu/boetticher/courses.htmlto see all the DBMS videos
+Gary Boetticher here is the correct link sce.uhcl.edu/boetticher/courses.html
Thanks i appreciate. Yeah, i was wondering why the link he gave me wasn't working.
Wouldn't this procedure not work if there is an odd amount of attributes in R?
+HimikoWerckmeister Even/Odd number of attributes is not an issue.
+Gary Boetticher Thank you for the reply, another pressing question I have is lets suppose I have the set of functional dependencies as follows:
BG->CD;
G->F;
CD->GH;
C->FG
F->D
I know for the fact that by deploying your LMR method as described in previous videos that the key can be either BG or BE, applying the algorithm described in the video I obtain the following decomposition G(G,F,D) and Rest(B,C,H) my question to you is, can I decompose these relations any further?
Why not use The Chase?
thats what I figured as well
4.49 that was scary...i thought samara (the RING) will appear :p
At 5:46, is C+ rather not CD? Thank you
+Gilbert Adu-Sefah Yes, I think it is. Because when R - BC = A+ = ACD (he included the D here. From the same logic C+ should have been CD). However even if C+ = CD, I do not think that this has an effect on the overall result in this problem.
I think he meant Z or Y, poor clarity indeed