MadeEasy Test Series: Databases - Database Normalization

Question

Comments

Answer should be c definitely in 2NF but transitive dependency may present.

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Counter example ...not in 2NF and only single candidate key.

R(ABCD)

AB->CD

B->C. (Partial dependency)

Candidate key:->AB

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yeah right sorry i thought single attribute candidate key.

Answer 1

A 3NF table that does not have multiple overlapping candidate keys is guaranteed to be in BCNF.

As here there is only single candidate key means there is no multiple overlapping candidate keys so if this is in 3NF then it is compulsory in bcnf.

U may see example here https://en.m.wikipedia.org/wiki/Boyce%E2%80%93Codd_normal_form

So B is correct ans.

Comments

@Pankaj  @srestha @Akriti @cse23

Plz verify me!!

Here question interpretation is required as well some English grammar.

Option A & C:->False bcz

If R has only one candidate key ,then R is in 2NF.(option A&C 's antecedent part says that it is compulsory in 2NF which need not be..see below counter example)

 

R(ABCD)

AB->CD

B->C. (Partial dependency)

Candidate key:->AB

So this relation is Not in 2NF.

-> if option C would [If R is in 2NF,then may not be in 3NF ;then it is abs. correct.]

Here option B is correct..Plz see my ans(though some intelli has downvoted it ;p)

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i think option B is correct cuz i could not find any counter example that if a relation is in 3NF and it has only one candidate key then it is not in BCNF..and clearly A and C are false as you explained.

but is it also true that if a relation has more than 1 candidate key that too overlapping aNd if it is in 3NF then may or not may be it is in BCNF.

see below example

R(A,B,C,D,E)

 

bc - > ade

cd  - > b

//here it is in 3NF as (bc,cd) are keys that too overlapping and it is also in BCNF.

aM I RIGHT??

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Yes ; U are correct.

And I suggest U to Visit my ans. Links below portion.

"3NF table not meeting BCNF (Boyce–Codd normal form)"

Answer 2

Let a relation R(A,B,C,D)  now AB is a ck . so fds can be A->C  , B->D  now they are not BCNF,3NF,2NF so option d looks right.

But b option specifically says if R is in 3NF so we have to take fds that are in 3NF only. Like AB->CD which is 3NF now it is also in BCNF . we cannot make any fd here with ab as ck which is 3nf but not bcnf

so option d is correct.