Tuesday, 11 July 2017

Find the key of relation R

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 the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = {AB C, A DE, B F, F GH, D IJ}. Find the key of relation R.

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. We can start with single LHS attributes.
  • 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
Description
A+
= ADE from the functional dependency A DE (By reflexivity rule)
= ADEIJ from D → IJ (By transitivity rule. A → D and D → IJ hence A → IJ)
No more functional dependencies that has either of the attributes ADEIJ in LHS. Hence, A+ = ADEIJ.
A+ ≠ R.
So, A is not a candidate key.
B+
= BF from B → F (By reflexivity rule)
= BFGH from F → GH (By transitivity rule)
No more functional dependencies that has either of the attributes BFGH in LHS. Hence, B+ = BFGH.
B+ ≠ R.
So, B cannot be a candidate key.



(AB)+
= ABC from AB C
= ABCDE from the functional dependency A DE
= ABCDEIJ from D → IJ
= ABCDEFIJ from B → F
= ABCDEFGHIJ from F → GH
(AB)+ = R.
So, (AB) is the candidate key.
**************


Go to - 1NF,    2NF,    3NF,    BCNF





 


Find the keys of a relational table in dbms solved problem




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