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 (Pseudo-transitive)	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 (Pseudo-transitive)	(CD)⁺ is equivalent to R. Hence CD forms a candidate key.
(AC)⁺	AC → D AD → B or CD → B	= ACD (Reflexive) = ABCD (Pseudo-transitive)	(AC)⁺ = R hence is a candidate key.

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

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find all the candidate keys of a relation for normalization purpose