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Recent questions tagged normalization

0 votes
2 answers 7 views
Consider relation R with attributes x, y, z and set of FD’s as, xy ->z ; z->x The highest normal form of R is? A)1NF B)2NF C) 3NF D) BCNF
asked 3 days ago in Databases Dhananjay15 14 points 7 views
0 votes
0 answers 4 views
If anyone having any idea how to solve mainly the (b) part. Please help me out. Consider a relation R with five attributes ABCDE.For each of the following instances of R, state whether it violates (a) the FD BC→ D and (b) the MVD BC →→ D: (a) { } (i.e., empty relation) (b) {(a,2,3,4,5), (2,a,3,5,5)} (c) {(a,2, ... , (a,2,3,6,5), (a,2,3,6,6)} (g) {(a,2,3,4,5), (a,2,3,6,5), (a,2,3,6,6), (a,2,3,4,6)}
asked Sep 13 in Databases कुशाग्र गुप्ता 46 points 4 views
0 votes
0 answers 4 views
Consider the following collection of relations and dependencies. Assume that each relation is obtained through decomposition from a relation with attributes ABCDEFGHI and that all the known dependencies over relation ABCDEFGHI are listed for each question. (The questions are independent of each other, obviously, since the given dependencies ... G, G → H 4. R4(D,C,H,G), A → I, I → A 5. R5(A,I,C,E)
asked Sep 11 in Databases कुशाग्र गुप्ता 46 points 4 views
1 vote
1 answer 40 views
When it comes to inserting something in a foreign key table, we cant put null values. But when it comes to deleting something from referenced table, we can perform ON DELETE SET NULL operation and make the foreign key value in a particular tuple as NULL. So, a foreign key can’t remain null while inserting but it can be null while deletion. Is my approach correct ?
asked Aug 21 in Databases user2525 1.5k points 40 views
0 votes
0 answers 12 views
Regarding statement Q, I don’t think it will violate 3NF because whatever be the R.H.S. if L.H.S. is superkey, then it is in BCNF and ultimately in 3NF. Therefore, it is allowed in 3NF and so the statement is supposed to be right. Is my approach correct ?
asked Aug 21 in Databases user2525 1.5k points 12 views
0 votes
0 answers 9 views In the above question : FDs are AB→C,AC→B,AD→E,B→D,BC→A,E→G. R1 ( ABC ) : AB → C AC → B BC → A R2 ( ACDE ) : AD → E R3 ( ADG ) has empty FDs. So, the above decomposition is not dependency preserving. But ... is AD a key order to check lossless decomposition ? Do we have 2 consider the closure of original fds to prove AD as key for R3 ( ADG ) ?
asked Aug 20 in Databases user2525 1.5k points 9 views
0 votes
0 answers 14 views
I know that BCNF may not satisfy dependency preserving but it should be lossless. So, when we are decomposing a relation for satisfying BCNF, we don’t care about the dependency preservation condition. But what about 3NF ? Is it compulsory to check for dependency preservation condition after a table has been decomposed to satisfy 3NF ?
asked Aug 20 in Databases user2525 1.5k points 14 views
0 votes
0 answers 20 views
Consider the following FDs : Sid → Sname Sname → Age The above set of FD is not - non trivial. I knw da definition of non - trivial in FD is that there should be no common element between X and Y attribute sets in the FD X → Y. The above set of FD is not non ... X and Y have something in common that is Sname. That is why the set of FDs is not non-trivial. Is my perception towards the FDs correct ?
asked Aug 19 in Databases user2525 1.5k points 20 views
0 votes
1 answer 22 views
0 votes
3 answers 32 views
Kindly show your solution. I am not able to decompose the relation. Thank you.
asked Aug 3 in Databases DukeThunders 132 points 32 views
0 votes
0 answers 3 views
Consider the following table the answer is there are two MVD in this relation Department ->>Part Department ->>Job how does the second MVD hold? d3 doesn’t have multiple values for job.
asked Jul 30 in Databases aditi19 29 points 3 views
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