Normalization
September 2, 2024Normalization
September 2, 2024Normalization
| 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?
| 3 | |
| 4 | |
| 5 | |
| 6 |
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.
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.
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.