NIELIT 2017 DEC Scientist B - Section B: 55

Question

Consider the relational schema $\text{R(A B C D)}$ with following functional dependency set $F=\{A\rightarrow BC,C\rightarrow D\};$ The relation $\text{R}$ is in 

• A. $2$NF
• B. BCNF
• C. $3$NF
• D. $1$NF

Comments

We can also go with this approach that we go A as candidate key , as A is the single attribute super key and it is also minimal so it is also candidate key and as it is single attribute so there will be no subsets of candidate key which will determine any non prime attribute or we can say that there will be no non-prime attribute which will be partially dependent an candidate key. So it is in 2NF.

Answer 1

here A is key....

So A->BC is in BCNF...

C->D ...is not in BCNF as C is not a super key ...its also not in 3NF as D is not a prime atribute...but its in 2nf as C is not a part of Key..

so overall relation is in 2NF

answer is option A

Answer 2

(A)2NF

R(A B C D)     HERE A is the CK

so for A->BC  A is the SUPER KEY so  it is in BCNF

BUT C->D    (NP->NP) so TRANSITIVE DEPENDENCY EXIST SO IT IS NOT IN 3NF

SO it is in 2NF

Answer 3

Given, functional dependencies are F={A→ BC, C→ D}
Finding candidate key of given relation is relation is
A+={A,B,C} because we don’t have an A element in right hand side.
A→ BC dependency A is the candidate key and also part of FD. So, it is lossless.
C→ D is following Transitive dependency. So, it never become 3NF.
According to given choices, A is the correct choice.

Answer 4

ANS : A  2NF

R(A B C D)     HERE A is the Candidate key

so for A→BC  A is the SUPER KEY so  it is in BCNF

BUT C→D    (NP->NP) so TRANSITIVE DEPENDENCY EXIST SO IT IS NOT IN 3NF

SO it is in 2NF

Answer 5

A relation is Not in 3NF if and only if there exists a non-trivial FD where Non-superkey determines a Non-prime attribute. 

Here, $C \rightarrow D$ violates 3NF because $C$ is Non-superkey & it determines $D$ which is non-prime attribute. So, the relation is Not in 3NF.

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A relation is Not in 2NF if and only if there exists a non-trivial FD where proper subset of a candidate key determines a Non-prime attribute. 

Since, the only candidate key here is $A,$ which doesn't have any proper subset, So, there is No violation of 2NF. Hence, $R$ is in 2NF.

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Answer 6

A is the key
now
for 3NF and FD X-->Y such that either  x is super key or Y is prime
if all fd follow such condition implies table to be in 3nf
here C-->D doesnot follow hence not in 3nf
 

for 2nf
if subset of ck finding non prime attribute then it violated 2nf but here no such
or saying if u have not any composite ck then u have always 2nf