Normalization

Question 27

Relation R has eight attributes ABCDEFGH . Fields of R contain only atomic values.

F = {CH→G, A→BC, B→CFG, E→A, F→EG}, is a set of functional dependencies (FDs) so that F⁺ is exactly the set of FDs that hold for R.

How many candidate keys does the relation R have?

A	3
B	4
C	5
D	6

Database-Management-SystemNormalizationGATE 2013

Question 27 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.
 Hence no. of candidate keys are 4: AD, BD, ED, FD.

Correct Answer: B

Question 27 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.
 Hence no. of candidate keys are 4: AD, BD, ED, FD.

Normalization

Normalization

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