CSJM University 2016 Question

Question

Consider a relation R(A,B,C,D,E) and FD's that hold on R are
A --> BC
CD --> E
B --> D
E --> A
Find the Normal form of R.

When I solved this I got CK's : A,E,BC,CD
Hence, Prime Attributes : A,B,C,D,E and No Non-Prime Attributes.

Comments

Highest normal form would be 3nf..

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when all attributes are prime then 3NF is sure. since B-> D not follow BCNF condition so not in BCNF. am Assuming you find keys correct,.

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3NF.

Answer 1

Here we have to think only about the FD

             B--->D

2NF

Here B and D are part of the candidate key BC and CD. So B and D become a Prime Attribute.

Since B and D are Prime Attribute, So it is not showing Partial Dependency.

Hence it is in 2NF.

3NF

Since  B is not the Super Key but D is the Prime Attribute, So it is in 3NF.

BCNF

Since B is not the Super Key.So it is not in BCNF.