Database-Management-System

Question 473

Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is __________ .

A	in 1NF but not in 2NF
B	in 2NF but not in 3NF
C	in 3NF but not in BCNF
D	in BCNF

Database-Management-SystemNormalizationUGC NET CS 2017 Nov- paper-3

Question 473 Explanation:

The attribute D is not part of any FD’s. So, D can be a candidate key or it may be part of the candidate key.
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+= {ABCDEFGH}
But CD+, GD+ and HD+ does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.

Correct Answer: A

Question 473 Explanation:

The attribute D is not part of any FD’s. So, D can be a candidate key or it may be part of the candidate key.
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+= {ABCDEFGH}
But CD+, GD+ and HD+ does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.