Find the candidate keys from the given relation for normalization
Question:
Consider
a relation with schema R(A,B,C,D) with functional dependencies (FD’s):
BC
→ A,
AD → B, CD → B, AC → D.
Find
all the candidate keys of R.
Solution:
Let us find the closure
for the left hand side (LHS) attributes of given set of functional
dependencies.
Let us take the FD BC → A to check whether
the LHS BC forms a candidate key or not.
We know

Result =

How?

Description

BC

BC

Given


BC

BCA

BC → A

If we know BC, then we can derive A uniquely as
per the reflexivity rule, hence result = BCA

BCA

BCAD

AC → D

From the previous step we know attribute A, and
by the FD AC → D the result
becomes ABCD which is equal to R. Hence, BC is a candidate key

We derived all the other candidate
keys in the same way as stated above and given in the table below;
LHS
Closure

Due to the FDs

Result becomes

Description

(BC)^{+}

BC → A
AC → D

=
ABC (Reflexive)
=
ABCD (Pseudotransitive)

The
result of (BC)^{+} is equivalent to R.
Hence
BC is a candidate key.

(AD)^{+}

AD → B

=
ABD (Reflexive)

(AD)^{+}
≠ R.
Hence
AD does not form
candidate key.

(CD)^{+}

CD → B
BC → A

= BCD (Reflexive)
=
ABCD (Pseudotransitive)

(CD)^{+}
is equivalent to R.
Hence
CD forms a candidate key.

(AC)^{+}

AC → D
AD → B
or
CD → B

=
ACD (Reflexive)
=
ABCD (Pseudotransitive)

(AC)^{+}
= R hence is
a candidate key.

BC,
AC, and CD are the candidate keys for the given relation.
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