Find the candidate keys of a table in dbms an example

Find the candidate keys of a relation, How to find the candidate keys, Which is the key for the given table, concept of candidate key in dbms, candidate key examples

Question:

Consider a relation R with attributes ABCDEFGH and functional dependencies S as follows:

S = {A → CD, ACF → G, AD → BEF, BCG → D, CF → AH, CH → G, D → B, H → DEG}

Find all keys for R.

Solution:

We can find keys (candidate keys) for a relation by finding the closure of an/set of attributes. Checking each attribute or all subsets of the given set of attributes for a key is time consuming job. Hence, we may employ some of the following heuristics/assumptions in identifying the keys;

• We may start checking all the left hand side attributes of any/all of the given set of functional dependencies.

• If we find the closure of an attribute and that attribute is the candidate key then any superset cannot be the candidate key. For example, if A is a candidate key, then AB is not a candidate key but a super key.

LHS	Result	Decision
A+	= ACD from A → CD = ACDBEF from AD → BEF = ACDBEFG from ACF → G = ACDBEFGH from CF → AH	Result includes all the attributes of relation R. That is, if we know A, then all the attributes of R could be uniquely determined. Hence, A is one candidate key.
ACF+	No need to find the closure of (ACF) because the subset A is already a candidate key.	ACF is a super key but not candidate key.
AD+	No need to find the closure of (AD) because the subset A is already a candidate key.	AD is a super key but not candidate key.
BCG+	= BCGD from BCG → D	Result does not include all the attribute of relation R. Hence, (BCG) cannot be a candidate key.
CF+	= CFAH from CF → AH Further, as we know A now, then we can conclude that CF will uniquely determine all the other attributes of A.	Result includes all the attributes of relation R. Hence, (CF) is one candidate key.
D+	= DB from D → B	Result does not include all R. Hence, D cannot be a key.
H+	= HDEG from H → DEG = HDEGB from D → B	Result does not include all R. Hence, H cannot be a key.

From the above table, it is clear that only A and CF are the candidate keys.

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How to find the super keys and candidate keys for normalization purpose in dbms