Advanced Database Management System - Tutorials and Notes: Find the candidate keys of a table in dbms an example

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Tuesday, 14 February 2017

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.



Go to - 1NF,    2NF,    3NF,    BCNF





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







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